1 GOY AL BROTHER S PR AKASH A N COORDINATE GEOMETRY A. SUMMA TIVE ASSESSMENT 7.1 DIST ANCE FORMULA 7 1. In the rectangular coordinate system, two number lines are drawn at right angles to each other. The point of intersection of these two number lines is called the origin whose coordinates are taken as (0, 0). The horizontal number line is known as the x-axis and the vertical one as the y-axis. 2. In the ordered pair (p, q), p is called the x-coordinate or abscissa and q is known as y-coordinate or ordinate of the point. 3. The coordinate plane is divided into four quadrants. 4. For a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively. The abscissa of a point is its perpendicular distance from y-axis. The ordinate of a point is its perpendicular distance from x-axis. 5. The abscissa of every point situated on the right side of y-axis is positive and the abscissa of every point situated on the left side of y-axis is negative. 6. The ordinate of every point situated above x-axis is positive and that of every point below x-axis is negative. 7. The abscissa of every point on y-axis is zero. 8. The ordinate of every point on x-axis is zero. 9. The distance between any two points P(x 1 , y 1 ) and Q (x 2 , y 2 ) is given by PQ = ( ( 2 2 2 1 2 1 – + – x x y y or PQ = ( ( 2 2 1 2 1 2 – + – x x y y PQ = ( ( 2 2 Difference of abscissae + Difference of ordinates 10. If O(0, 0) is the origin and P(x, y) is any point, then from the above formula, we have : OP = ( ( 2 2 2 2 –0 + –0 = + x y x y 11. Three points are said to be collinear if they are on the same straight line. 12. For three points to be collinear, the sum of the distances between two pairs of points is equal to the distance between the third pair of points. TEXTBOOK’S EXERCISE 7.1 Q.1. Find the distance between the following pairs of points : (i) (2, 3), (4, 1) (ii) (–5, 7), (–1, 3) (iii) (a, b), (–a,–b) Sol. We know that distance between two points (x 1 , y 1 ) and (x 2 , y 2 ) = ( ( 2 2 2 1 2 1 – + – x x y y (i) Distance between the points (2, 3) and (4, 1) Question Bank In Mathematics Class X (Term–II)
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COORDINATE GEOMETRY
A. SUMMATIVE ASSESSMENT
7.1 DISTANCE FORMULA
7
1. In the rectangular coordinate system, twonumber lines are drawn at right angles to eachother.
The point of intersection of these twonumber lines is called the origin whosecoordinates are taken as (0, 0). The horizontalnumber line is known as the x-axis and thevertical one as the y-axis.
2. In the ordered pair (p, q), p is called the
x-coordinate or abscissa and q is known as
y-coordinate or ordinate of the point.
3. The coordinate plane is divided into four
quadrants.
4. For a given point, the abscissa andordinate are the distances of the given point fromy-axis and x-axis respectively.
The abscissa of a point is its perpendiculardistance from y-axis.
The ordinate of a point is its perpendiculardistance from x-axis.
5. The abscissa of every point situated on theright side of y-axis is positive and the abscissa ofevery point situated on the left side of y-axis isnegative.
6. The ordinate of every point situated abovex-axis is positive and that of every point belowx-axis is negative.
7. The abscissa of every point on y-axis is zero.
8. The ordinate of every point on x-axis is zero.
9. The distance between any two pointsP(x1, y1) and Q (x2, y2) is given by
PQ = ( ) ( )2 22 1 2 1– + –x x y y
or PQ = ( ) ( )2 21 2 1 2– + –x x y y
PQ =( )
( )
2
2
Difference of abscissae
+ Difference of ordinates
10. If O(0, 0) is the origin and P(x, y) is anypoint, then from the above formula, we have :
OP = ( ) ( )2 2 2 2– 0 + – 0 = +x y x y
11. Three points are said to be collinear ifthey are on the same straight line.
12. For three points to be collinear, the sum ofthe distances between two pairs of points is equalto the distance between the third pair of points.
TEXTBOOK’S EXERCISE 7.1
Q.1. Find the distance between the followingpairs of points :
(i) (2, 3), (4, 1) (ii) (–5, 7), (–1, 3)
(iii) (a, b), (–a, –b)
Sol. We know that distance between two points
(x1, y1) and (x2, y2)
= ( ) ( )2 22 1 2 1– + –x x y y
(i) Distance between the points (2, 3) and (4, 1)
Question Bank In Mathematics Class X (Term–II)
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= ( ) ( )2 24 – 2 + 1 – 3 = 4 + 4 = 8 = 2 2
(ii) Distance between the points (–5, 7) and(–1, 3)
= ( ){ } ( ){ }2 2–1 – –5 + 3 – 7
= 16 +16 = 32 = 4 2(iii) Distance between the points (a, b) and
(–a, –b)
= ( ) ( )2 2– – + – –a a b b = ( ) ( )2 2
–2 + –2a b
= 2 24 + 4a b = ( )2 2 2 24 + = 2 +a b a b
Q.2. Find the distance between the points(0, 0) and (36, 15). [Imp.]
Sol. The given points are (0, 0) and (36, 15)
Distance = ( ) ( )2 236 – 0 + 15 – 0
[Using distance formula]
= ( ) ( )2 236 + 15 = 1296 + 225
= 1521 = 3 × 3 × 13 × 13 = 39
Q.3. Determine if the points (1, 5), (2, 3) and(–2, –11) are collinear. [Imp.]
Sol. The given points are A(1, 5), B(2, 3),
C(–2, –11)We know that, distance between two points
d = ( ) ( )2 22 1 2 1– + –x x y y
AB = ( ) ( )2 22 –1 + 3 – 5
= 1+ 4 = 5 … (i)
BC = ( ) ( )2 2–2 – 2 + –11 – 3
= ( ) ( )2 2– 4 + –14
= 16 +196 = 212 … (ii)
CA = { } { }2 21 – (–2) + 5 – (–11)
= 2 2(3) + (16)
= 9 + 256 = 265 … (iii)
From equation (i), equation (ii) and equation (iii),We have
AB + BC CABC + CA ABCA + AB BC
Hence, the given points are not collinear.
Q.4. Check whether (5, –2), (6, 4) and(7, –2) are the vertices of an isosceles triangle.
Sol. The given points are A(5, –2), B(6, 4) and
C(7, –2)We know that, distance between two points
d = ( ) ( )2 22 1 2 1– + –x x y y
AB = ( ) ( ){ }226 – 5 + 4 – –2
= ( ) ( )2 21 + 6 = 1 + 36 = 37 ... (i)
BC = ( ) ( )2 27 – 6 + –2 – 4
= ( ) ( )2 21 + –6 = 1 + 36 = 37 … (ii)
From equation (i) and equation (ii), AB = BC
Hence, given points are the vertices of anisosceles triangle.
Q.5. In a classroom, 4 friends are seated atthe points A, B, C and D as shown in figure.Champa and Chameli walk into the class andafter observing for a few minutes Champa asksChameli,
‘Don't you think ABCD is a square?’’Chameli disagrees. Using distance formula, findwhich of them is correct.
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
ARows
Columns
B
C
D
Sol. From the figure, the coordinates ofpoints are A(3, 4), B(6, 7), C(9, 4) and D(6, 1)respectively.
We know that distance between two points (x1, y1)and (x2, y2)
= ( ) ( )2 22 1 2 1– + –x x y y
AB = ( ) ( )2 26 – 3 + 7 – 4 = ( ) ( )2 2
3 + 3
= 9 + 9 = 18 = 3 2 … (i)
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BC = ( ) ( )2 29 – 6 + 4 – 7 = ( ) ( )2 2
3 + –3
= 9 + 9 = 18 = 3 2 … (ii)
CD = ( ) ( )2 26 – 9 + 1 – 4 = ( ) ( )2 2
–3 + –3
= 9 + 9 = 18 = 3 2 … (iii)
DA = ( ) ( )2 23 – 6 + 4 – 1 = ( ) ( )2 2
–3 + 3
= 9 + 9 = 18 = 3 2 … (iv)
AC = ( ) ( )2 29 – 3 + 4 – 4 = 6 … (v)
BD = ( ) ( )2 26 – 6 + 1 – 7 = 6 … (vi)
From equations (i), (ii), (iii) and (iv), we have
AB = BC = CD = DA i.e., all the four sides areequal
Also, from (v) and (vi), we have AC = BD
i.e., the diagonals are equal
ABCD is a square.
Hence, Champa is correct
Q.6. Name the type of quadrilateral formed,if any, by the following points, and give reasonsfor your answer :
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Sol. We know that distance between two points
(x1, y1) and (x2, y2)
= ( ) ( )2 22 1 2 1– + –x x y y
(i) The given points are A(–1, –2), B(1, 0),C(–1, 2), D(–3, 0)
AB = ( ){ } ( ){ }2 21 – –1 + 0 – –2
= ( ) ( )2 22 + 2
= 4 + 4 = 8 = 2 2 … (1)
BC = ( ) ( )2 2–1 –1 + 2 – 0
= ( ) ( )2 2–2 + 2 = 4 + 4
= 8 = 2 2 … (2)
CD = ( ){ } { }2 2–1– – 3 + 2 – 0
= { } { }2 2–1+ 3 + 2 = ( ) ( )2 2
2 + 2
= 4 + 4 = 8 = 2 2 … (3)
AD = ( ){ } ( ){ }2 2–3 – –1 + 0 – –2
= ( ) ( )2 2–3 +1 + 2 = ( ) ( )2 2
–2 + 2
= 4 + 4 = 8 = 2 2 … (4)
AC = ( ){ } ( ){ }2 2–1 – –1 + 2 – –2
= ( ) ( )2 2–1+1 + 4 = 16 = 4 … (5)
BD = ( ){ } ( )2 21 – –3 + 0 – 0
= ( )24 = 16 = 4 … (6)
From equation (1), (2), (3), (4), (5) and (6), allsides of the quadrilateral ABCD are equal anddiagonals AC and BD are equal. Hence, quadrilateralABCD is a square.
(ii) The given points are A(–3, 5), B(3, 1),C(0, 3), D(–1, – 4)
AB = ( ) ( )2 23 + 3 + 1 – 5 = 36 + 16
= 52 = 2 13 … (1)
BC = ( ) ( )2 20 – 3 + 3 – 1
= 9 + 4 = 13 … (2)
CD = ( ) ( ) ( ) ( )2 2 2 20 +1 + 3 + 4 = 1 + 7
= 1+ 49 = 50 = 5 2 … (3)
DA = ( ){ } ( )2 2–1 – –3 + – 4 – 5
= ( ) ( )2 2–1+ 3 + –9 = ( ) ( )2 2
2 + –9
= 4 + 81 = 85 … (4)
AC = ( ){ } { }2 20 – –3 + 3 – 5
= ( ) ( )2 23 + –2 = 9 + 4 = 13 … (5)
DB 2 23 1 1 4
16 25 41 …(6)
From equation (1), (2), (3), (4), (5) and (6) we seethat BC + AC = AB
A, B, C are collinear.
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So, it is not possible to draw ABC.
Hence, ABCD does not form any quadrilateral.
(iii) Given points are A(4, 5), B(7, 6), C(4, 3),D(1, 2)
AB = ( ) ( ) ( ) ( )2 2 2 27 – 4 + 6 – 5 = 3 + 1
= 9 + 1 = 10 … (1)
BC = ( ) ( )2 24 – 7 + 3 – 6
= ( ) ( )2 2–3 + –3 = 9 + 9
= 18 = 3 2 … (2)
CD = ( ) ( )2 21 – 4 + 2 – 3
= ( ) ( )2 2–3 + –1 = 9 + 1 = 10 … (3)
DA = ( ) ( )2 21 – 4 + 2 – 5 = 9 + 9
= 18 = 3 2 … (4)
Thus from equation (1), (2), (3) and (4). It is clearthat opposite sides are equal.
Hence, ABCD is a parallelogram.
Q.7. Find the point on the x-axis which isequidistant from (2, –5) and (–2, 9).
Sol. We know that a point on the x-axis is of the
form (x, 0), where y coordinate is always zero. Letpoint P(x, 0) be equidistant from A(2, –5) and B(–2, 9).
Thus PA = PB
We know, d2 = (x2 – x1)2 + (y2 – y1)
2
PA2 = PB2
(2 – x)2 + (–5 – 0)2 = (–2 – x)2 + (9 – 0)2
4 + x2– 4x + 25 = 4 + x2 + 4x + 81
–8x = 81 – 25, –8x = 56
x =56
–8= –7
Hence, the required point is (–7, 0).
Q.8. Find the values of y for which thedistance between the points P(2, –3) and Q(10, y)is 10 units. [Imp.]
Sol. We have, PQ = 10 [Given]
PQ2 = 102 = 100 [Squaring both sides]
(10 – 2)2 + {y – (–3)}2 = 100
[Distance, d = ( ) ( )2 22 1 2 1– + –x x y y ]
(8)2 + (y + 3)2 = 100
64 + y2 + 6y + 9 = 100
y2 + 6y – 27 = 0
y2 + 9y – 3y – 27 = 0
y(y + 9) – 3(y + 9) = 0
(y + 9) (y – 3) = 0 y = – 9 or y = 3
Hence, the required value of y is –9 or 3.
Q.9. If Q(0, 1) is equidistant from P(5, –3)and R(x, 6), find the value of x. Also find thedistances QR and PR. [V. Imp.]
Sol. Given points are Q(0, 1), P(5, –3) and R(x, 6).
PQ = QR
( ) ( )2 25 – 0 + –3 –1 = ( ) ( )2 2
– 0 + 6 –1x
( ) ( )2 2225 + – 4 = + 5x
25 + 16 = x2 + 25
x2 = 16 x = ± 4
Thus, coordinates of point R are (±4, 6)
QR = ( ) ( )2 20 ± 4 + 1 – 6
= 16 + 25 = 41If x = – 4,
PR = ( ) ( ){ }22– 4 – 5 + 6 – –3
= 81+ 81 = 9 2If x = 4,
PR = ( ) ( ){ }224 – 5 + 6 – –3 = 1+ 81 = 82
Q.10. Find a relation between x and y suchthat the point (x, y) is equidistant from the points(3, 6) and (–3, 4). [HOTS]
Sol. Let the point P(x, y) is equidistant from the
points A(3, 6) and B(–3, 4)
As per condition, PA = PB
PA2 = PB2
(3 – x)2 + (6 – y)2 = (–3 – x)2 + (4 – y)2
9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x
+ 16 + y2 – 8y
x2 + y2 – 6x – 12y + 45 = x2 + y2 + 6x
– 8y + 25
–12x – 4y + 20 = 0 12x + 4y – 20 = 0
3x + y – 5 = 0
Hence, relation between x and y is 3x + y – 5 = 0.
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OTHER IMPORTANT QUESTIONS
Q.1. If the distance between the points(2, –2) and (–1, x) is 5, one of the values of x is :
(a) (9, –2) (b) 2 (c) –1 (d) 1
Sol. (b) We have 5 = ( ) ( )2 22 +1 + –2 – x
25 = 9 + 4 + x2 + 4x x2 + 4x – 12 = 0
x2 + 6x – 2x – 12 = 0
x(x + 6) – 2(x + 6) = 0
(x + 6) (x – 2) = 0 x = 2, or x = – 6
Q.2. The distance of the point P(2, 3) fromthe x-axis is :
(a) 2 (b) 3 (c) 1 (d) 5
Sol. (b) The distance of P(2, 3) from x-axis is its
y-coordinate i.e., 3.
Q.3. The distance of the point P(– 6, 8) fromthe origin is :
Q.19. Find the value of x, if the distancebetween the points (x, – 1) and (3, –2) is x + 5.
[2011 (T-II)]
Sol. The given points are A(x, – 1) and B(3, –2)
Then, x1 = x, y1 = –1 and x2 = 3, y2 = –2
AB = 2 22 1 2 1( ) ( )x x y y
x + 5 = 2 2(3 ) ( 2 1)x
x + 5 = 29 6 1x x
x2 + 25 + 10x = x2 – 6x + 10
16x = –15 x =15
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Q.20. Show that the points A(5, 6), B(1, 5),C(2, 1) and D(6, 2) are the vertices of a square.
[2004]
Sol. AB = ( ) ( )2 21 – 5 + 5 – 6
= 16 + 1 = 17
BC = ( ) ( )2 22 –1 + 1 – 5
= 1 + 16 = 17
CD = ( ) ( )2 26 – 2 + 2 –1
= 16 + 1 = 17
DA = ( ) ( )2 25 – 6 + 6 – 2
= 1 + 16 = 17 AB = BC = CD = AD
Now, diagonal
AC = ( ) ( )2 22 – 5 + 1 – 6
= 9 + 25 = 34and diagonal
BD = ( ) ( )2 26 –1 + 2 – 5
= 25 + 9 = 34So, diagonal AC = diagonal BD.
Since ABCD is a quadrilateral whose all the sidesare equal and both the diagonals are equal. Thus,quadrilateral ABCD is a square. Proved.
Q.21. Prove that the points (0, 0), (5, 5) and(–5, 5) are the vertices of a right isoscelestriangle. [2005, 2011 (T-II)]
Sol. AB = ( ) ( )2 25 – 0 + 5 – 0
= 25 + 25 = 50 = 5 2
BC = ( ) ( )2 2– 5 – 5 + 5 – 5
= ( )2–10 + 0 = 100 = 10
AC = ( ) ( )2 2– 5 – 0 + 5 – 0 = 25 + 25
= 50 = 5 2
AB2 + AC2 = ( ) ( )2 25 2 + 5 2
= 50 + 50 = 100 = BC2
By converse of Pythagoras theorem, we concludethat ABC is a right isosceles triangle, right angled atA. Proved.
Q.22. A circle has its centre at the origin anda point P(5, 0) lies on it. The point Q(6, 8) liesoutside the circle. Is it true? Justify your answer.
[HOTS]
Sol. Radius OP = ( ) ( )2 20 – 5 + 0 – 0
= 25 = 5 units
Also OQ = ( ) ( )2 20 – 6 + 0 – 8
= 36 + 64 = 10 units
OQ > OP, hence, Q lies outside the circle.
Hence, the statement is true.
Q.23. Find a point which is equidistant fromthe points A(–5, 4) and B(–1, 6). How many suchpoints are there? [HOTS]
Sol. Let the required point be P(x, y).
Then, PA = PB PA2 = PB2
(x + 5)2 + (y – 4)2 = (x + 1)2 + (y – 6)2
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x2 + 25 + 10x + y2 + 16 – 8y
= x2 + 1 + 2x + y2 + 36 – 12y
10x – 8y + 41 = 2x – 12y + 37
8x + 4y + 4 = 0 2x + y + 1 = 0 … (i)
By hit and trial method, one solution of (i) is
x = –1, y = 1.
Hence, (–1, 1) is the required point. Aslo, infinitenumber of points are there which are equidistant fromthe given points. In fact, all points which are solutionsof the equation 2x + y + 1 = 0 are equidistant from thegiven points.
Q.24. The points A(2, 9), B(a, 5) and C(5, 5)are the vertices of a triangle ABC right angled atB. Find the value of a and hence the area ofABC.
Sol. We have AC2 = AB2 + BC2
[Pythagoras theorem] (2 – 5)2 + (9 – 5)2
= (a – 2)2 + (5 – 9)2 + (a – 5)2 + (5 – 5)2
25 = a2 + 4 – 4a + 16 + a2 + 25 – 10a
2a2 – 14a + 20 = 0 a2 – 7a + 10 = 0
a2 – 5a – 2a + 10 = 0
a(a – 5) – 2 (a – 5) = 0
(a – 5) (a – 2) = 0 a = 5 or 2
a = 2 [Rejecting a = 5, because in this case Bcoincides with C]
Area of the triangle =1
2× AB × BC
= ( ) ( )
( ) ( )
2 2
2 2
1× 2 – 2 + 9 – 5
2
× 5 – 2 + 5 – 5
=1
2× 4 × 3 = 6 sq. units.
Q.25. Find the circumcentre of the triangle,whose vertices are (–2, –3), (–1, 0) and (7, –6).
[HOTS]
Sol. Let the coordinates of the circumcentre of
the triangle be P(x, y). Then, we know thatcircumcentre of a triangle is equidistant from each ofits vertices.
PA = ( ) ( )2 2+ 2 + + 3x y
PA2 = x2 + y2 + 4x + 6y + 13 …(i)
PB = ( ) ( )2 2+ 1 + – 0x y
PB2 = x2 + y2 + 2x + 1 …(ii)
PC = ( ) ( )2 2– 7 + + 6x y
PC2 = x2 + y2 – 14x + 12y + 85 …(iii)
Now, PA = PB
PA2 = PB2
x2 + y2 + 4x + 6y + 13 = x2 + y2 + 2x + 1
2x + 6y = – 12 x + 3y = – 6 …(iv)
And PB = PC PB2 = PC2
x2 + y2 + 2x + 1 = x2 + y2 – 14x + 12y + 85
16x – 12y = 84 4x – 3y = 21 …(v)
Solving (iv) and (v), we get x = 3, y = – 3.
Hence, circumcentre of the triangle is (3, –3).
Q.26. If the point (x, y) is equidistant fromthe points (a + b, b – a) and (a – b, a + b), provethat bx = ay. [V. Imp.]
Sol. Let P(x, y), Q(a + b, b – a) and R(a – b,
a + b) be given points.
Then, PQ = PR
( ){ } ( ){ }2 2– + + – –x a b y b a
{x – (a + b)}2 + {y – (b – a)}2
= {x – (a – b)}2 + {y – (a + b)}2
x2 – 2x (a + b) + (a + b)2 + y2 – 2y(b – a) + (b – a)2 = x2 + (a – b)2
– 2x (a – b) + y2 – 2y (a + b) + (a + b)2
– 2x (a + b) – 2y (b – a)= – 2x (a – b) – 2y (a + b)
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ax + bx + by – ay = ax – bx + ay + by
2bx = 2ay bx = ay. Proved.
Q.27. The two opposite vertices of a squareare (1, –6) and (5, 4). Find the coordinates of theother two vertices. [HOTS]
Sol. Let ABCD be the given square whose two
opposite vertices are A(1, –6) and C(5, 4). Let B(x, y)be its third vertex.
Then, AB = BC AB2 = BC2
(x – 1)2 + (y + 6)2 = (x – 5)2 + (y – 4)2
x2 + y2 – 2x + 12y + 37 = x2 + y2 – 10x
– 8y + 41
8x + 20y – 4 = 0 2x + 5y = 1
y =1 – 2
5
x … (i)
From right triangle ABC, we have,
AB2 + BC2 = AC2
(x – 1)2 + (y + 6)2 + (x – 5)2 + (y – 4)2
= (5 – 1)2 + (4 + 6)2
(x2 + y2 – 2x + 12y) + (x2 + y2 – 10x – 8y)
+ 78 = 116
2(x2 + y2 – 6x + 2y) = 38
x2 + y2 – 6x + 2y = 19
x2 +( )2 2 1 – 21 – 2
– 6 +5 5
xxx
= 19
[Using (i)]
25x2 + (1 + 4x2 – 4x) – 150x + 10 – 20x = 475
29x2 – 174x – 464 = 0
x2 – 6x – 16 = 0 (x – 8) (x + 2) = 0
x = – 2 or x = 8
Now, x = – 2 y = 1 and x = 8
y = – 3
Hence, the remaining vertices are B(–2, 1) andD(8, – 3) or B(8, – 3) and D(–2, 1).
Q.28. Find the coordinates of the pointequidistant from three given points A(5, 3),B(5, –5) and C(1, –5). [2006]
Sol. Let P(x, y) be the required point. As P(x, y)
is equidistant from A(5, 3), B(5, –5) and C(1, –5), so,AP = BP = CP.
Case I. AP = BP
( ) ( )2 2– 5 + – 3x y = ( ) ( )2 2
– 5 + + 5x y
2 2–10 + 25 + – 6 + 9x x y y
= 2 2–10 + 25 + + 10 + 25x x y y
x2 + y2 – 10x – 6y + 34
= x2 + y2 – 10x + 10y + 50
– 6y – 10y + 34 – 50 = 0
– 16y = 16 y = – 1. ….. (i)
Case II. BP = CP
( ) ( )2 2– 5 + + 5x y = ( ) ( )2 2
–1 + + 5x y
2 2–10 + 25 + + 10 + 25x x y y
= 2 2– 2 + 1 + + 10 + 25x x y y
x2 + y2 – 10x + 10y + 50
= x2 + y2 – 2x + 10y + 26
– 8x = 26 – 50 = – 24
x = 3. …(ii)
The coordinates of the point equidistant from A,B and C are P(3, – 1).
Q.29. Find the coordinates of the centre of acircle passing through the points A(2, 1),B(5, –8) and C(2, – 9). Also, find the radius ofthis circle.
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Sol. Let P(x, y) be the centre of the circle passing
through the points A(2, 1), B(5, – 8) and C(2, –9).
Then PA = PB = PC
PA2 = PB2 = PC2
Now, PA2 = PB2
(x – 2)2 + (y – 1)2 = (x – 5)2 + (y + 8)2
x2 + y2 – 4x – 2y + 5 = x2 + y2 – 10x
+ 16y + 89
6x – 18y – 84 = 0
x – 3y = 14 …(i)
And, PB2 = PC2
(x – 5)2 + (y + 8)2 = (x – 2)2 + (y + 9)2
x2 + y2 – 10x + 16y + 89
= x2 + y2 – 4x + 18y + 85
6x + 2y – 4 = 0 3x + y = 2 … (ii)
On solving (i) and (ii), we get x = 2 and y = – 4
Hence, the coordinates of the centre of the circleare P (2, – 4).
Radius of the circle = PA (PA = PB = PC)
= ( ) ( )2 22 – 2 + 1 + 4 = 2 20 + 5
= 25 = 5 units.
Q.30. Find the coordinates of thecircumcentre of a triangle whose vertices areA(4, 6), B(0, 4) and C(6, 2). Also, find itscircumradius. [HOTS]
Sol. Let A(4, 6), B(0, 4) and C(6, 2) be the
vertices of the given triangle ABC.
Let P(x, y) be the circumcentre of ABC.
Then, PA = PB = PC
PA2 = PB2 = PC2
Now, PA2 = PB2
(x – 4)2 + (y – 6)2 = (x – 0)2 + (y – 4)2
x2 + y2 – 8x – 12y + 52
= x2 + y2 – 8y + 16
8x + 4y = 36 2x + y = 9 …(i)
Again, PB2 = PC2
(x – 0)2 + (y – 4)2 = (x – 6)2 + (y – 2)2
x2 + y2 – 8y + 16 = x2 + y2 – 12x – 4y + 40
12x – 4y = 24 3x – y = 6 … (ii)
On solving (i) and (ii), we have x = 3 and y = 3.
Coordinates of the circumcentre of ABC areP(3, 3)
Circumradius = PA = ( ) ( )2 24 – 3 + 6 – 3
= 2 21 + 3 = 10 units.
Q.31. If P(2, –1), Q(3, 4), R(–2, 3) andS(–3, –2) be four points in a plane, show thatPQRS is a rhombus but not a square. Find thearea of the rhombus. [2006, 2011 (T-II)]
Sol. Let, P(2, –1), Q(3, 4), R(–2, 3) and
S(–3, –2) be the vertices of a quadrilateral. Join PRand QS.
Now, PQ = ( ) ( )2 23 – 2 + 4 + 1
= 2 21 + 5 = 26 units
QR = ( ) ( )2 2–2 – 3 + 3 – 4
= ( ) ( )2 2–5 + –1 = 26 units
RS = ( ) ( )2 2–3 + 2 + –2 – 3
= ( ) ( )2 2–1 + –5 = 26 units
and SP = ( ) ( )2 22 + 3 + –1 + 2
= 2 25 +1 = 26 units
PQ =QR = RS = SP = 26 units.
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ANDiagonal PR = ( ) ( )2 2
–2 – 2 + 3 +1
= ( ) ( )2 2– 4 + 4
= 32 = 4 2 units.
Diagonal QS = ( ) ( )2 2–3 – 3 + –2 – 4
= ( ) ( )2 2– 6 + – 6
= 72 = 6 2 units.
diagonal PR diagonal QS.
Thus, PQRS is a quadrilateral having all sidesequal but diagonals unequal.
Hence, PQRS is a rhombus, but not a square.Proved.
Area of the rhombus PQRS
=1
2× product of diagonals
=1
× PR × QS2
= ( )1× 4 2 × 6 2
2sq. units = 24 sq. units.
Q.32. If two vertices of an equilateral
triangle be (0, 0), ( )3, 3 , find the third vertex.
[V. Imp.]
Sol. O(0, 0) and A ( )3, 3 be the given points
and let B(x, y) be the third vertex of equilateralOAB.
Then, OA = OB = AB
OA2 = OB2 = AB2
We have, OA2 = (3 – 0)2 + ( )23 – 0 = 12
OB2 = x2 + y2
and AB2 = (x – 3)2 + ( )2– 3y
AB2 = x2 + y2 – 6x – 2 3 y + 12
Now, OA2 = OB2 and OB2 = AB2
x2 + y2 = 12 and, x2 + y2
= x2 + y2 – 6x – 2 3 y + 12
x2 + y2 = 12 and 6x + 2 3 y = 12
x2 + y2= 12 and 3x + 3 y = 6
x2 +
26 – 3
3
x = 12
3 + 3 = 6
6 – 3=
3
x y
xy
3x2 + (6 – 3x)2 = 36 12x2 – 36x = 0 x = 0 or x = 3
x = 06
3 = 6 = = 2 33
y y
and x = 3 9 + 3 = 6y
6 – 9
= = – 33
y
Hence, the coordinates of the third vertex B are
( ) ( )0, 2 3 or 3, – 3 .
Q.33. If P and Q are two points whose
coordinates are (at2, 2at) and 2
2,
a a
tt
respectively and S is the points (a, 0), show that
1 1+
SP SQis independent of t. [V. Imp.]
Sol. We have,
SP = ( ) ( )2 22 – + 2 – 0at a at
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= ( )22 2–1 + 4a t t = a(t2 + 1)
and SQ =
2 2
2
2– + – 0
a aa
tt
=( )22 2 2
4 2
1 – 4+
a t a
t t
SQ = ( )22 22
1 – + 4a
t tt
= ( )222
1 +a
tt
= ( )22
1 +a
tt
1 1
+SP SQ
= ( ) ( )2
2 2
1+
+ 1 1 +
t
a t a t
= ( )2
2
1 + 1=
+ 1
t
aa t,
which is independent of t. Proved.
Q.34. Two vertices of an isosceles triangleare (2, 0) and (2, 5). Find the third vertex if thelength of the equal sides is 3.
Sol. Let the coordinates of A be (x, y).
Then AB2 = (3)2 = (2 – x)2 + (0 – y)2
4 – 4x + x2 + y2 = 9
x2 + y2 = 5 + 4x ….. (i)
Again AC2 = (3)2 = ( ) ( )2 22 – + 5 –x y
4 – 4x + x2 + 25 – 10y + y2 = 9
29 – 4x – 10y + y2 + x2 = 9
x2 + y2 – 10y – 4x = 9 – 29
x2 + y2 – 4x – 10y = – 20 ….. (ii)
Putting the value of x2 + y2 from (i) in (ii),we have
5 + 4 – 10 – 4x y x = – 20
– 10y = – 20 – 5 = – 25
y = 2.5 =5
2….. (iii)
Substituting, y = 2.5 in (i), we have
x2 + y2 = 5 + 4x
x2 +25
4= 5 + 4x
4x2 + 25 = 20 + 16x 4x2 – 16x + 5 = 0 …... (iv)
x =( )2
16 ± 16 – 4 × 4 × 5
2× 4[Using quadratic formula]
=16 ± 256 – 80
8=
16 ± 176 16 ± 4 11=
8 8
x =4 ± 11
2= 2 ±
11
2
coordinates of the third vertex are
11 5 11 5A 2 – , , 2 + ,
2 2 2 2
Q.35. Prove that the points (2a, 4a), (2a, 6a),
and 2 + 3 , 5a a a are the vertices of an
equilateral triangle. [2011 (T-II)]
Sol. Let A(2a, 4a), B(2a, 6a) and C(2a + 3 a,
5a) be the vertices of a triangle.
AB = ( ) ( )2 22 – 2 + 6 – 4a a a a
= ( )2 22 = 4a a = 2a
AC = 2 2
2 + 3 – 2 + 5 – 4a a a a a
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= 2 23 +a a = 24a = 2a.
BC 2 2(2 3 2 ) (5 6 )a a a a a
2 2 23 4 2a a a a
Thus, AB = AC = BC
ABC is an equilateral triangle. Proved.
Q.36. If (– 4, 3) and (4, 3) are two vertices ofan equilateral triangle, find the coordinates ofthe third vertex, given that the origin lies in theinterior of the triangle. [HOTS]
Sol. AB = 2 2– 4 – 4 – 3 – 3 = 8 units.
Clearly, the x-axis cuts AB at its mid-point.
Also, O is in the interior of ABC.
So, the third vertex will lie on the y-axis.
[ y-axis the perpendicular bisector of AB soany point on the y-axis is equidistant from A and B]
Let the coordinates of vertex C be (0, –y).
AC = AB AC2 = AB2
Then,
(4 – 0)2 + (3 + y)2 = 64
16 + 9 + y2 + 6y = 64 y2 + 6y – 39 = 0
PRACTICE EXERCISE 7.1A
Choose the correct option (Q.1 – 8) :1. The distance between the points A(0, 6)
and B(0, –2) is :(a) 6 (b) 8 (c) 4 (d) 2
2. The distance between the points (0, 5) and(–5, 0) is :
(a) 5 (b) 5 2 (c) 2 5 (d) 10
3. Three points A, B, C are said be collinear,if :
(a) they lie on the same straight line
(b) they do not lie on the same straight line
(c) they lie on three different straight lines
(d) none of these
4. The distance between the points P(2, –3)and Q(2, 2) is :
(a) 2 units (b) 3 units
(c) 4 units (d) 5 units
5. The points M(0, 6), N(–5, 3) and P(3, 1)are the vertices of a triangle, which is :
(a) isosceles (b) equilateral
(c) scalene (d) right angled
6. The coordinates of the vertices of an
equilateral triangle are A(3, y), B(3, 3 ) and
C(0, 0). The value of y is :(a) 4 (b) 5 (c) –1 (d) none of these
7. If the distance between the points (4, c),and (1, 0) is 5, then c is :
(a) ± 4 (b) 4 (c) – 4 (d) 0
8. The distance between the points P(a cos35°, 0) and Q(0, a cos 65°) is :
(a) a (b) 2a
(c) 3a (d) none of these
9. Points A(3, 1), B(12, –2) and C(0, 2)cannot be the vertices of a triangle. Is it true?
[Imp.]
y =– 6 ± 36 + 4×1 × 39
2
=– 6 ± 192 – 6 ± 8 3
=2 2
= –3 ± 4 3
y = –3 – 4 3 or –3 + 4 3
y = –3 + 4 3 [Rejecting y = –3 – 4 3 ,as it will give the coordinates of C above x-axis]
Hence, coordinates of the third vertex are (0, –y)
or (0, 3 – 4 3 )
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10. Name the type of triangle formed by thepoints A(–5, 6), B(– 4 –2) and C(7, 5).
11. Find the value of x if the distancebetween the points A(–3, –14) and B(x, –5) is 9units. [Imp.]
12. If the point P(2, – 4) is equidistant fromX(3, 8) and Y(–10, y), find the values of y. Aslo,find the distance XY. [Imp.]
13. Find the distance between the followingpair of points :
(i) (a sin , – b cos ) and(– a cos , b sin )
(ii) (a + b, b + c) and (a – b), (c – b)
(iii) (a, 0) and (0, b)
14. Show that the points A(a, b + c),B(b, c + a), and C(c, a + b) are collinear. [Imp.]
15. Find all possible values of a for whichthe distance between the points A(a, –1) andB(5, 3) is 5 units. [Imp.]
16. Find the point on x-axis which isequidistant from (–2, 5) and (2, –3).
17. Show that the points P(–3, 2) Q(–5, 5)R(2, 3) and S(4, 4) are the vertices of a rhombus.Find the area of the rhombus. [Imp.]
18. Prove that the points A(3, 4), B(– 4, 3),C(5, 0) lie on the circle with centre O(0, 0).
19. Show that the following points are thevertices of isosceles triangles :
(i) (0, 6), (–5, 3), (3, 1)
(ii) (0, 0), (a, b), (–a, b).
20. The length of a line segment is 10. If oneend is at (2, –3), and the abscissa of the second
end is 10, find its ordinate.
21. Find the circumcentre of the trianglewhose vertices are (0, –3), (7, 0) and (4, 7).
22. Find the coordinates of the circumcentreof the triangle whose vertices are (8, 6), (8, –2)and (2, –2). Also, find its circumradius. [V. Imp.]
23. If the points (x, y) is equidistant from thepoints A(5, 1) and B(–1, 5), prove that3x = 2y. [2005]
24. If the distance P(x, y) from the pointsA(3, 6) and B(–3, 4) are equal, prove that 3x +y = 5. [2008]
25. Show that A(–3, 2), B(–5, –5), C(2, –3)and D(4, 4) are the vertices of a rhombus. [2008]
26. The centre of a circle is (2 – 1, 7) andit passes through the point (–3, –1). If thediameter of the circle is 20 units, then find thevalue of . [2009]
27. Find the points on the x-axis which are at
a distance of 2 5 from the point (7, – 4). How
many such points are there? [2011 (T-II)]
28. Find those points on the x-axis which areat a distance of 5 units from the point (5, –3).
[2011 (T-II)]
29. Find the value of 's' if the point P(0, 2)is equidistant from Q(3, s) and R(s, 5).
[2011 (T-II)]
30. A point P is at a distance of 10 from
the point (2, 3). Find the coordinates of the pointP if its y coordinate is twice its x coordinate.
[2011 (T-II)]
7.2 SECTION FORMULA
1. The coordinates of the point P(x, y) whichdivides the line segment joining A(x1, y1) andB(x2, y2) internally in the ratio m : n, are given by :
2 2 2 2+ += , =
+ +
mx nx my nyx y
m n m n
2. The coordinates of the mid-point M of aline segment AB with end points A(x1, y1) and
B(x2, y2) are :1 2 1 2+ +
,2 2
x x y y
3. The point of intersection of the medians oftriangle is called its centroid.
4. The coordinates of the centroid of thetriangle whose vertices are (x1, y1), (x2, y2) and
(x3, y3) are given by1 2 3 1 2 3+ + + +
,3 3
x x x y y y
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Q.1. Find the coordinates of the point whichdivides the join of (–1, 7) and (4, –3) in the ratio2 : 3. [Imp.]
Sol. Let the coordinates of the point which divide
the join of points A(–1, 7) and B(4, –3) in the ratio2 : 3 be (x, y).
Using the section formula, we get
x =2 1+
+
mx nx
m n=
( )( ) ( )( )2 4 + 3 –1
2 + 3
=8 – 3 5
=5 5
= 1
And, y =2 1+
+
my ny
m n=
( )( ) ( )( )2 –3 + 3 7
2 + 3
=– 6 + 21 15
=5 5
= 3
Hence, the co-ordinates of required point are (1, 3).
Q.2. Find the coordinates of the points oftrisection of the line segment joining (4, –1) and(–2, –3). [Imp.]
Sol. The given points are A(4, –1) and B(–2, –3).
Let the points of trisection of the line segment ABbe P(x1, y1) and Q(x2, y2) respectively.
Then AP = PQ = QB
Thus, P divides AB in the ratio 1 : 2 internally andQ divides AB in the ratio 2 : 1 internally.
Using section formula, we get
x1 =( ) ( ) ( ) ( )1 –2 + 2 4
1+ 2=
–2 + 8 6=
3 3= 2
And, y1 =( ) ( ) ( )( )1 –3 + 2 –1
1+ 2=
– 3 – 2 5= –
3 3
Hence, the coordinates of point P are–5
2,3
.
Again, x2 =( )( ) ( )( )2 –2 + 1 4
2 +1=
– 4 + 4
3= 0
And, y2 =( )( ) ( )( )2 –3 + 1 –1
2 +1=
– 6 – 1 7= –
3 3
Hence, the coordinates of point Q are7
0, –3
.
Q.3. To conduct Sports Day activities, inyour rectangular shaped school ground ABCD,lines have be drawn with chalk powder at adistance of 1 m each. 100 flower pots have beenplaced at a distance of 1 m from each other
along AD, as shown in figure. Niharika runs1
4th
the distance AD on the 2nd line and posts a green
flag. Preet runs1
5th the distance AD on the
eighth line and posts a red flag. What is thedistance between both the flags? If Rashmi has topost a blue flag exactly halfway between the linesegment joining the two flags, where should shepost her flag?
Sol. Taking A as origin, AB as x-axis and AD as
y-axis. The distance covered by Niharika.
=1
4distance of AD on the 2nd line
=1
4× 100 = 25 m
Thus, coordinates of green flag are (2, 25)
Similarly, the position of red flag
= Distance covered by Preet on the eighth line
=1
5distance AD on 8th line
=1
5× 100 = 20 m
TEXTBOOK’S EXERCISE 7.2
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Thus, coordinates of red flag are (8, 20).
Using distance formula the distance betweengreen and red flag is
= ( ) ( )2 28 – 2 + 20 – 25 = 36 + 25 = 61 m
Position of blue flag = Mid-point of green flagand red flag
By section formula co-ordinates of blue flag
=2 + 8 25 + 20
,2 2
= (5, 22.5)
Hence, blue flag is in 5th line at a distance of22.5 m along AD.
Q.4. Find the ratio in which the line segmentjoining the points (–3, 10) and (6, –8) is dividedby (–1, 6). [V. Imp.]
Sol. Let A(–3, 10), B(6, –8) and P(–1, 6)
Let P divide AB in the ratio K : 1.
–1 =( ) ( ) ( )( )K 6 + 1 –3
K +1
6K – 3
K +1= –1 6K – 3 = – K – 1 7K = 2
K =2
7and 6 =
( )( ) ( )( )K –8 + 1 10
K +1
–8K + 10
K +1= 6 –8K + 10 = 6K + 6
14K = 4 K =4 2
=14 7
Hence, the required ratio is 2 : 7.
Q.5. Find the ratio in which the line segmentjoining A(1, –5) and B(– 4, 5) is divided by thex-axis. Also find the coordinates of the point ofdivision.
Sol. Given A(1, –5) and B(– 4, 5)
Let the point of division be P. Let the ratio be K : 1.
Q.6. If (1, 2), (4, y), (x, 6) and (3, 5) are thevertices of a parallelogram taken in order, find xand y. [2011 (T-II)]
Sol. Let ABCD be a parallelogram. Given
vertices of a parallelogram ABCD are A(1, 2), B(4, y),C(x, 6) and D(3, 5). We know that diagonals of aparallelogram bisect each other. Let E is the mid-pointof diagonal AC.
Coordinates of E are+1 6 + 2
,2 2
x
or+1
, 42
x … (i)
Also, E is the mid-point of diagonal BD.
Coordinates of E are3 + 4 5 +
,2 2
y
or7 5 +
,2 2
y …. (ii)
From equation (i) and equation (ii), we have
+1, 4
2
x =
7 5 +,
2 2
y
Comparing both sides, we have,+1
2
x=
7
2
x + 1 = 7 x = 6 And5 +
2
y= 4
5 + y = 8 y = 3
Hence, the values of x and y are 6 and 3respectively.
Q.7. Find the coordinates of a point A, whereAB is the diameter of a circle whose centre is(2, –3) and B is (1, 4). [2011 (T-II)]
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Sol. Let O (2, –3) be the centre of the circle.
Let the coordinates of the point A be (x, y)
AB is a diameter of a circle whose centre is O.
O is the mid-point of AB.
So, by section formula+1 + 4
,2 2
x y
= (2, –3)
+1
2
x= 2 x + 1 = 4 x = 3
And+ 4
2
y= –3 y + 4 = –6 y = –10
Hence, the coordinates of the point A are(3, –10).
Q.8. If A and B are (–2, –2) and (2, – 4)respectively, find the coordinates of P such that
AP =3
7AB and P lies on the line segment AB.
[2011 (T-II)]
Sol. We have AP =3
7AB
7AP = 3AB 7AP = 3(AP + PB)
[ P lies on the line segment AB]
7AP = 3AP + 3PB
7AP – 3AP = 3PB 4AP = 3PB
AP
PB=
3
4
Let the coordinate of P be (x, y).
Then, x =( )( ) ( )( )3 2 + 4 –2
3 + 4=
6 – 8 2= –
7 7
y =( )( ) ( )( )3 –4 + 4 –2
3 + 4=
–12 – 8 20= –
7 7
Hence, the coordinates of the point P are
2 20– , –
7 7
.
Q.9. Find the coordinates of the points whichdivide the line segment joining A(–2, 2) andB(2, 8) into four equal parts. [2011 (T-II)]
Sol. Let P(x1, y1), Q(x2, y2) and R(x3, y3) be the
points which divide the line segment AB into fourequal parts i.e. AP = PQ = QR = RB
So, P divides AB in the ratio 1 : 3 internally.
x1 =( )( ) ( )( )1 2 + 3 –2
1+ 3=
2 – 6 4= –
4 4= –1
And y1 =( )( ) ( )( )1 8 + 3 2
1+ 3=
8 + 6 14 7= =
4 4 2
Thus, coordinates of P are7
–1,2
.
Also, Q divides AB in the ratio 1 : 1, i.e., Q is themid-point of AB.
x2 =–2 + 2
2= 0 and y2 =
2 + 8 10=
2 2= 5
Thus, coordinates of Q are (0, 5), R divides AB inthe ratio 3 : 1.
x3 =( )( ) ( )( )3 2 + 1 –2 6 – 2 4
= =3 +1 4 4
= 1
And y3 =( )( ) ( )( )3 8 + 1 2 24 + 2 26 13
= = =3 +1 4 4 2
Thus, coordinates of R are13
1,2
.
Q.10. Find the area of a rhombus if itsvertices are (3, 0), (4, 5), (–1, 4) and (–2, –1)taken in order. [2011 (T-II)]
Sol. Let ABCD be a rhombus whose vertices are
A(3, 0), B(4, 5), C(–1, 4) and D(–2, –1). AC and BDare diagonals.
We know that the area of a rhombus
=1
2(Product of its diagonals)
Area of the rhombus ABCD =1
2(AC × BD)
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=1
2( ) ( )2 2–1 – 3 + 4 – 0
× ( ) ( )2 2– 2 – 4 + –1 – 5
=1
2( ) ( )16 + 16 × 36 + 36
=1
2( ) ( )32 × 72 =
1
2( ) ( )4 2 × 6 2
=1
2× 24 × 2 = 24 square units.
OTHER IMPORTANT QUESTIONS
Q.1. If the centroid of the triangle formed bythe points (a, b), (b, c) and (c, a) is at the origin,then a3 + b3 + c3 is equal to :
(a) abc (b) 0 (c) a + b + c (d) 3abc
Sol. (d) We have+ +
3
a b c= 0 and
+ +
3
b c a= 0
a + b + c = 0
Now, a3 + b3 + c3 = 3abc, if a + b + c = 0
Q.2. The line segment joining points (–3, – 4)and (1, – 2) is divided by y-axis in the ratio :
(a) 1 : 3 (b) 2 : 3 (c) 3 : 1 (d) 2 : 3
Sol. (c) Let the required ratio be K : 1 and the
coordinates of the point of intersection be (0, y).
Then, 0 = K ×1 –1× –3
K +1K = 3
Hence, required ratio is 3 : 1.
Q.3. The ratio in which (4, 5) divides the joinof (2, 3) and (7, 8) is :
(a) 4 : 3 (b) 5 : 2
(c) 3 : 2 (d) 2 : 3
Sol.(d) Let the required ratio be K : 1.
Then, 4 =K × 7 +1× 2
K +14k + 4 = 7k + 2
3K = 2 K =2
3
Required ratio is2
3: 1 = 2 : 3
Q.4. The x-axis divides the join of A(2, –3)and B(5, 6) in the ratio :
(a) 1 : 2 (b) 2 : 1 (c) 3 : 5 (d) 2 : 3
Sol. (a) Let the required ratio be k : 1 and the
coordinates of the point of intersection be (x, 0).
Then, 0 =( )× 6 +1× –3
+1
k
k k =
1
2
Required ratio is 1 : 2.
Q.5. The point which divides the linesegment joining the point (7, –6) and (3, 4) in theratio 1 : 2 internally lies in the :
(a) I quadrant (b) II quadrant
(c) III quadrant (d) IV quadrant
Sol. (d) Let the coordinates of the required point
be (x, y).
Then x =1×3 + 2× 7 17
=3 3
and y =( )1× 4 + 2× –6 –8
=3 3
Coordinates of the point are17 –8
,3 3
.
Clearly, this point lies in IV quadrant.
Q.6. If P , 43
a is the mid-point of the line
segment joining the points Q(– 6, 5) andR(–2, 3), then the value of a is :
(a) – 4 (b) –12 (c) 12 (d) – 6
Sol. (b) We have–6 – 2
=3 2
aa = –12
Q.7. A line intersects the x-axis and the y-axis at the points P and Q respectively. If (2, –5)is the mid point of PQ, then the coordinates ofpoints P and Q are; respectively.
(a) (4, 0), (0, –10) (b) (0, 10), (– 4, 0)
(c) (0, 4), (–10, 0) (d) (0, –10), (4, 0)
Sol. (a) Let the line intersects the x-axis at (x, 0)
and the y-axis at (0, y).
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Then,+ 0
2
x= 2 and
0 +
2
y= –5
x = 4 and y = –10
Coordinates of P and Q are (4, 0) and (0, –10)respectively.
Q.8. Find the coordinates of a point onx-axis which divides the line segment joining thepoints (–2, –3) and (1, 6) in the ratio 1 : 2.
[2011 (T-II)]
Sol. Let the coordinates of the point which divide
the join of points A(–2, –3) and B(1, 6) in the ratio
1 : 2 be (x, y).
Using the section formula, we get
x =
2 1 (1)(1) 2( 2)
1 2
mx nx
m n=
1 4 3
3 3= –1
And, y =2 1 (1)(6) 2( 3)
1 2
my ny
m n
+ + −=
+ +
=
6 6 0
3 3= 0
Hence, the coordinates of required point are
(–1, 0).
Q.9. Find the ratio in which the point
2 20,
7 7divides the join of (–2, –2) and
(2, – 4). [2011 (T-II)]
Sol. Let the point C divide AB in the ratio : 1.
Then, the coordinates of C are
2 2 4 2,
1 1
But, the coordinates of C are given as2 20
,7 7
− −
2 2 2
1 7and
4 2 20
1 7
14 – 14 = –2 – 2 and –28 – 14 = –20 – 20
16 = 12 and –8 = –6 =3
4Hence, the point C divides AB in the ratio 3 : 4.
Q.10. Find the coordinates of a point Rwhich divides the line segment joining the points
P(–2, 3) and Q(4, 7) internally in the ratio4
7.
[Imp.]
Sol. Let the coordinates of the required point be
(x, y). Then
x =( )4× 4 + 7× –2 16 –14 2
= =4 + 7 11 11
And, y =4× 7 + 7× 3 49
=4 + 7 11
Hence, coordinates of the required point are
2 49,
11 11
.
Q.11. Find the ratio in which the point
– 6, 4
5
divides the join of the points
(3, 5) and (– 4, 5). [Imp.]Sol.
Let the required ratio be k : 1.
Then,–6
5=
3 – 4
+1
k
k
– 6k – 6 = 15 – 20k 14k = 15 + 6
k =21 3
=14 2
required ratio is 3 : 2.
Q.12. Find the coordinates of the pointswhich divide the line segment joining the points(–2, 0) and (0, 8) in four equal parts. [2005C]
Sol. Let the points P, Q, R divide the join of
A(–2, 0) and B(0, 8) in four equal parts. Then
Q is the mid point of AB, so, coordinates of or are
–2 + 0 0 + 8,
2 2
i.e., Q(–1, 4).
P is the mid-point of AQ, so, coordinates of P are
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–2 –1 0 + 4,
2 2
i.e., P
–3,2
2
.
R is the mid point of QB, so, coordinates of Q are
–1+ 0 4 + 8,
2 2
i.e., R
–1,6
2
Hence, the required points are :
P3
– , 22
, Q(–1, 4), R
1– ,6
2
.
Q.13. In what ratio does the point P(2, –5)divide the line segment joining A(–3, 5) andB(4, –9)? [2005C]
Sol. Let the required ratio be k : 1.
Then, by the section formula, the coordinates of P
are :4 – 3 –9 + 5
,+1 +1
k k
k k
4 – 3
+1
k
k= 2 and
–9 + 5
+1
k
k= –5
4k – 3 = 2k + 2 and – 9k + 5 = – 5k – 5
2k = 5 and 4k = 10 k =5
2in each case.
Hence, the required ratio is5
2: 1, which is 5 : 2.
Q.14. Prove that (4, –1), (6, 0), (7, 2) and(5, 1) are the vertices of a rhombus. Is it asquare? [V. Imp.]
Sol. Let the given points be A, B, C and D
respectively. Then,
Coordinates of the mid-point of AC are :
4 + 7 –1+ 2 11 1, = ,
2 2 2 2
Coordinates of the mid-point of BD are :
6 + 5 0 +1 11 1, = ,
2 2 2 2
Thus, AC and BD have the same mid-points.
Hence, ABCD is a parallelogram.
Now,
AB = ( ) ( )2 26 – 4 + 0 +1 = 5
BC = ( ) ( )2 27 – 6 + 2 – 0 = 5
AB = BC
So, ABCD is a parallelogram, whose adjacentsides are equal.
Hence, ABCD is a rhombus. Proved.
We have,
AC = ( ) ( )2 27 – 4 + 2 +1 = 3 2 ,
and BD = ( ) ( )2 26 – 5 + 0 –1 = 2 .
Clearly, AC BD.
So, ABCD is not a square.
Q.15. If the mid-point of the line segmentjoining the points A(3, 4) and B(k, 6) is P(x, y)and + y – 10 = 0, find the value of k.
[2011 (T-II)]
Sol. We have x =3 +
2
kand y =
4 + 6
2
2x = 3 + k and y = 5
Also, x + y – 10 = 0 3 +
+ 5 – 102
k= 0
k = 7
Q.16. ABCD is a parallelogram withvertices A(x1, y1), B(x2, y2) and C(x3, y3). Findthe coordinates of the fourth vertex D in terms ofx1, x2, x3, y1, y2, y3.
Sol. Let the coordinates of D be (x, y).
Then, we know that the diagonals of aparallelogram bisect each other.
Therefore, mid-point of AC = mid-point of BD.
1 3 1 3+ +
,2 2
x x y y =
2 2+ +,
2 2
x x y y
1 3 2+ +
=2 2
x x x x
and1 3 2+ +
=2 2
y y y y
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x = x1 – x2 + x3, y = y1 – y2 + y3
Therefore, the coordinates of D are (x1 – x2 + x3,y1 – y2 + y3)
Q.17. If (9a – 2, –b) divides the line segmentjoining A(3a + 1, –3) and B(8a, 5) in the ratio3 : 1, find the values of a and b. [V. Imp.]
Sol. We have, 9a – 2 =( )3×8 +1× 3 +1
3 +1
a a
36a – 8 = 24a + 3a + 1
9a = 9 a = 1
And – b =3× 5 –1×3
3 +1– 4b = 12 b = –3
Q.18. Find the ratio in which the line2x + 3y – 5 = 0 divides the line segment joiningthe points (8, –9) and (2, 1). Also, find thecoordinates of the point of division. [2011 (T-II)]
Sol. Let the required ratio be k : 1 and the
coordinates of division are (x, y).
Then, x =2 + 8
+1
k
kandy =
– 9
+1
k
k…... (i)
But, P(x, y) lies on the line 2x + 3y – 5 = 0
2 + 8 – 9
2 . + 3 . – 5+1 +1
k k
k k= 0
4k + 16 + 3k – 27 – 5k – 5 = 0
2k = 16 k = 8 Required ratio = 8 : 1
Also, from (i) x =2 × 8 + 8
9=
8
3and y =
1–
9Hence, coordinates of the point of division are
8 1, –
3 9
.
Q.19. If the point C(–1, 2) divides internallythe line segment joining A(2, 5) and B(x, y) in theratio 3 : 4, then find the coordinates of B.
[2011 (T-II)]
Sol. Let the coordinates of B be (,). It is given
that AC : BC = 3 : 4.So, the coordinates of C are
3 4 2 3 4 5 3 8 3 20, ,
3 4 3 4 7 7
α + × β + × α + β +=
+ + But, the coordinates of C are (–1, 2).
3 8
7
α += –1 and
3 20
7= 2
= –5 and = –2Thus, the coordinates of B are (–5, –2).
Q.20. If the coordinates of the mid-points ofthe sides of a triangle are (1, 1), (2, –3) and(3, 4), find its centroid.
Sol. Let P(1, 1), Q(2, –3) and R(3, 4) be the mid-
points of sides AB, BC and CA respectively of triangleABC. Let A(x1, y1), B(x2, y2) and C(x3, y3) be thevertices of triangle ABC. Then, P is the mid-point ofAB.
1 2 1 2+ +
= 1, = 12 2
x x y y
x1 + x2 = 2 and y1 + y2 = 2 …(i)
Q is the mid-point of BC.
2 3 2 3+ +
= 2, = –32 2
x x y y
x2 + x3 = 4 and y2 + y3 = –6 …(ii)
R is the mid-point of AC
1 3+
= 32
x xand
1 3+= 4
2
y y
x1 + x3 = 6 and y1 + y3 = 8 ...(iii)
From (i), (ii) and (iii), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 6
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and y1 + y2 + y2 + y3 + y1 + y3 = 2 – 6 + 8
x1 + x2 + x3 = 6 and y1 + y2 + y3 = 2 ...(iv)
The coordinates of the centroid of ABC
are : 1 2 3 1 2 3+ + + +,
3 3
x x x y y y
.
=6 2 2
, = 2,3 3 3
Q.21. A(3, 2) and B(–2, 1) are two vertices
of a triangle ABC whose centroid G has the
coordinates 5 1, –
3 3
. Find the coordinate of the
third vertex C of the triangle. [2004]
Sol. Let the coordinates of the third vertex of
triangle ABC be C(x, y).
Then,3 + – 2 5
=3 3
x
+1 5=
3 3
x
x = 5 – 1 = 4
Again,2 +1+ 1
= –3 3
y
3 + 1= –
3 3
y
y = –1 – 3 = – 4
Coordinates of third vertex are C(4, – 4)
Q.22. If the points (10, 5), (8, 4) and (6, 6)are the mid-points of the sides of a triangle, findits vertices. [2006]
Sol. Let the vertices of the triangle be A(x1, y1),
B(x2, y2) and E(x3, y3).
(10, 5) is the mid-point of A(x1, y1) and B(x2, y2)
10 =1 2+
2
x xand 5 =
1 2+
2
y y
x1 + x2= 20 … (i)
and y1 + y2= 10 … (ii)
(8, 4) is the mid-point of B(x2, y2) and C(x3, y3).
x2 + x3= 16 … (iii)
and y2 + y3= 8 … (iv)
(6, 6) is the mid-point of C(x3, y3) and A(x1, y1)
x1 + x3 = 12 … (v)
and y1 + y3= 12 … (vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x3 = 24 … (vii)
Adding (ii), (iv) and (vii), we get
y1 + y2 + y3 = 15 … (viii)
Subtracting (i) from (vii), we get x3 = 4
Subtracting (ii) from (viii), we get y3 = 5
Similarly, we get x1 = 8 and y1 = 7, x2 = 12 and
y2 = 3
Hence, coordinates of the vertices are (8, 7), (12,
3) and (4, 5).
Q.23. If (4, –8), B(3, 6) and C(5, – 4) are thevertices of ABC, D is the mid-point of BC and
P is a point on AD joined such thatAP
PD= 2.
Find the coordinates of P. [2008]
Sol. Let the coordinates of P be (x, y).
Coordinates of D are– 4 + 65 + 3
,2 2
or (4, 1).
P divides AD is the ratio 2 : 1
x =2× 4 +1× 4
3and y =
( )2×1+1× –8
3
x = 4 and y = –2
Hence, coordinates of P are (4, –2).
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PRACTICE EXERCISE 7.2A
Q.24. If the points A(6, 1), B(8, 2), C(9, 4)and D(p, 3) are the vertices of a parallelogramtaken in order, find the value of p. [2011 (T-II)]
Sol. We know that the diagonals of a
parallelogram bisect each other. So, coordinates of themid-points of diagonal AC are same as the coordinatesof the mid-point of diagonal BD.
6 9 1 4 8 2 3, ,
2 2 2 2
p
15 5 8 5, ,
2 2 2 2
p
15 8
2 2
p15 = 8 + p p = 7
Q.25. The line segment joining the pointsA(2, 1) and B(5, –8) is trisected at the points P
and Q such that P is nearer to A. If P also lies onthe line given by 2x – y + k = 0, find the valueof k. [2009, 2011 (T-II)]
Sol. Since the line segment AB is trisected at
point P and Q.
So, AP = PQ = QB
AP : PB = 1 : 2
x coordinate of P is1×5 + 2× 2
1+ 2= 3
y coordinate of P is( )1× –8 + 2×1
1+ 2= –2
Coordinates of P are (3, –2)
Since, P lies on the line 2x – y + k = 0
2 × 3 + 2 + k = 0 k = –8.
Choose the correct option (Q. 1 – 8) :
1. If the ratio in which P divides the linesegment joining (x1, y1) and (x2, y2) be k : 1, thencoordinates of the point P are :
(a)1 2 2 1+ +
,+1 +1
kx x ky y
k k
(b)2 1 2 1+ +
,+1 +1
kx x ky y
k k
(c)1 2 1 2+ +
,+1 +1
x x y y
k k
(d) none of these
2. The y-axis divides the join of P(– 4, 2) andQ(8, 3) in the ratio :
(a) 3 : 1 (b) 1 : 3 (c) 2 : 1 (d) 1 : 2
3. Two vertices of PQR are P(–1, 4) andQ(5, 2) and its centroid is G(0, –3). Thecoordinates of R are :
(a) (4, 3) (b) (4, 15)
(c) (– 4, –15) (d) (–15, – 4)
4. A point A divides the join of X(5, –2) andY(9, 6) in the ratio 3 : 1. The coordinates of A
are :(a) (4, 7) (b) (8, 4)
(c)11
,52
(d) (12, 8)
5. The point M(1, 2) divides the join ofP(–2, 1) and Q(7, 4) in the ratio :
(a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 2 : 3
6. If M(–1, 1) is the mid point of the linesegment joining P(–3, y) and Q(1, y + 4), thenthe value of y is :
(a) 1 (b) –1 (c) 2 (d) 0
7. The coordinates of the centroid of thetriangle with vertices (a, 0), (0, b) and(a, b) are :
(a) ,2 2
a b (b) ,
3 3
a b
(c)2 2
,3 3
a b (d) none of these
8. If the point P(2, 1) lies on the line segmentjoining points A(4, 2) and B(8, 4), then
(a) AP =1
3AB (b) AP = PB
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(c) PB =1
3AB (d) AP =
1
2AB
9. In what ratio does the point P(2, –5)divide the line segment joining A(–3, 5) andB(4, –9)?
10. What are the coordinates of the centroidof a triangle whose vertices are (0, 6), (8, 12)and (8, 0)?
11. Find the coordinates of a point A, whereAB is the diameter of a circle whose centre is(2, –3) and B is (1, 4).
12. Check whether A(4, 3), B(6, 4), C(5, –6)and D(–3, 5) are the vertices of a parallelogram.
13. Point P(5, –3) is one of the two points oftrisection of the line segment joining the pointsA(7, –2) and B(1, –5). Is this statement true?
14. In what ratio does the x-axis divide theline segment joining the points (– 4, –6) and(–1, 7)? Find the coordinates of the point ofdivision.
15. If (a, b) is the mid-point of the linesegment joining the points A(10, –6) and B(k, 4),and a – 2b = 18, find the value of k and thedistance AB. [Imp.]
16. The line segment joining the pointsA(3, 2) and B(5, 1) is divided at the point P inthe ratio 1 : 2 and it lies on the line 3x – 18y+ k = 0. Find the value of k. [V. Imp.]
17. Find the coordinates of the point R onthe line segment joining the points P(–1, 3) and
Q(2, 5) such that PR =3
5PQ. [V. Imp.]
18. If the coordinates of the mid-points ofthe sides of a triangle are (1, 2), (0, –1) and(2, –1), find the coordinates of its vertices.
19. Find the length of the medians of aABC whose vertices are A(7, –3), B(5, 3) and
C(3, –1).20. If the mid-points of the sides of a triangle
are (2, 3),3
,42
and
11,5
2
, find the centroid
of the triangle.
21. Three vertices of a parallelogram are(a + b, a – b), (2a + b, 2a – b) and(a – b, a + b). Find the fourth vertex. [V. Imp.]
22. Prove that the coordinates of the centroidof a ABC with vertices A(x1, y1), B(x2, y2) andC(x3, y3) are given by
1 2 3 1 2 3+ + + +,
3 3
x x x y y y [2009]
23. If A(5, –1), B(–3, –2) and C(–1, 8) arethe vertices of a triangle ABC, find the length ofthe median through A and the coordinates of thecentroid. [2006]
24. The mid-points of the sides of a triangleare (3, 4), (4, 6) and (5, 7). Find the coordinatesof the vertices of the triangle. [2008]
25. The line segment joining the points P(3,3) and Q(6, –6) is trisected at the points A and Bsuch that A is nearer to P. If A also lies on theline given by 2x + y + k = 0, find the value of k.
[2009]
26. If P(x, y) is any point on the line segmentjoining the points A(a, 0) and B(0, b) then show
thatx y
a b+ = 1 [2011 (T-II)]
27. Using the section formula, show that thepoints A(–3, 1), B(1, 3) and C(–1, 1) arecollinear. [2011 (T-II)]
28. Prove that the parallelogramcircumscribing a circle is a rhombus.
[2011 (T-II)]
7.3 AREA OF A TRIANGLE1. The area of a ABC with vertices
A(x1, y1), B(x2, y2) and C(x3, y3) is given by :
Area ABC =
( ) ( ) ( ){ }1 2 3 2 3 1 2 1 2
1– + – + –
2x y y x y y x y y
Since area of a triangle cannot be negative,we consider the absolute or numerical value ofthe area.
2. Three given points A(x1, y1), B(x2, y2) andC(x3, y3), are collinear if area of ABC = 0
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1
2[x1 (y2 – y3) + x2(y3 – y1)
+ x3(y1 – y2)] = 0
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
TEXTBOOK’S EXERCISE 7.3
Q.1. Find the area of the triangle whosevertices are :
(i) (2, 3), (–1, 0), (2, – 4)
(ii) (–5, –1), (3, –5), (5, 2)
Sol. Area of the triangle with vertices (x1, y1),
(x2, y2) and (x3, y3)
=1
2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
(i) Vertices of trianlge are (2, 3), (–1, 0) and (2, – 4).
Area of triangle
=1
2[2{0 – (– 4)} + (–1)(– 4 –3)
+ 2(3 – 0)] sq. units
=1
2(8 + 7 + 6) =
21
2sq. units.
(ii) Vertices of triangle are (–5, –1), (3, –5) and(5, 2).
Area the triangle
=1
2[(–5) {–5 – 2} + 3 {2 – (–1)}
+ 5{(–1) – (–5)}] sq. units
=1
2[35 + 9 + 20] = 32 sq. units.
Q.2. In each of the following find the valueof k, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k) [Imp.]
(ii) (8, 1), (k, – 4), (2, –5) [Imp.]
Sol.(i) Let given points (7, –2), (5, 1) and (3, k)
are collinear.
Area of the triangle formed by these points
=1
2[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
=1
2[7 (1 – k) + 5{k – (–2)} + 3(–2 –1)]
=1
2[7 – 7k + 5k + 10 – 9]
=1
2[8 – 2k] = 4 – k
If the points are collinear, then area of the triangle= 0
4 – k = 0 k = 4
(ii) Let the given points (8, 1), (k, – 4) and(2, –5) are collinear.
Area of the triangle
=1
2[8{– 4 –(–5)} + k (–5)}
+ k (–5 –1) + 2{1 – (– 4)}]
=1
2[8 –6k + 10] =
1
2[18 –6k] = 9 – 3k
If the points are collinear, then area of the triangle= 0
9 – 3k = 0 3k = 9 = k = 3.
Q.3. Find the area of the triangle formed byjoining the mid-points of the sides of the trianglewhose vertices are (0, –1), (2, 1) and (0, 3). Findthe ratio of this area to the area of the giventriangle.
Sol. Let A, B, C be the vertices of the triangle
ABC. Let D, E and F be the mid-points of sides BC,CA and AB respectively. Then,
Coordinates of point D are2 + 0 1+ 3
,2 2
i.e., (1, 2).
Coordinates of point E are( )3 + –10 + 0
,2 2
i.e.,
(0, 1).
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Coordinates of point F are( )1+ –12 + 0
,2 2
i.e.,
(1, 0).
Area of the triangle DEF
=1
2[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
=1
2[1 (1 – 0) + 0 (0 – 2) + 1 (2 – 1)]
=1
2[1 + 0 + 1] = 1 sq. unit
Again, area of the triangle ABC
=1
2[0 (1 – 3) + 2 {3 – (–1)} + 0 (–1 – 1)]
= 4 sq. units.
Ratio of the area of the triangle formed byjoining mid-points to the area of the given triangle= 1 : 4.
Q.4. Find the area of the quadrilateralwhose vertices, taken in order, are (– 4, –2),(–3, –5), (3, –2) and (2, 3). [Imp.]
Sol. Let A(– 4, –2), B(–3, –5), C(3, –2) and
D(2, 3) are the vertices of the quadrilateral ABCD.
Since, area of the triangle with vertices (x1, y1),(x2, y2) and (x3, y3)
=1
2[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
Join BD.
So, area of ABD =1
2[(– 4){–5 – 3)}
+ (–3) {3 – (–2)} + 2{(–2) – (–5)}]
=1
2[32 – 15 + 6] =
23
2sq. units
Area of CBD
=1
2[3(–5 – 3) + (–3){3 – (–2)}
+ (2){(–2) – (–5)}]
=1
2[–24 – 15 + 6]
= –33
2=
33
2sq. units [numerically]
Area of the quadrilateral ABCD= Area of the triangle ABD+ Area of the triangle CBD
=23
2sq. units +
33
2sq. units = 28 sq. units
Q.5. You know that, a median of a triangledivides it into two triangles of equal areas. Verifythis result for ABC whose vertices are A(4, –6),B(3, –2) and C(5, 2). [HOTS]
Sol. Let D be the mid-point of the side BC of the
triangle ABC. Then
The coordinates of point D are( )–2 + 23 + 5
,2 2
= (4, 0)Then, AD is median
Now, area of ABD =1
2[4{(–2) – 0}
+ 3{0 – (–6)} + 4{(–6) – (–2)}]
=1
2[–8 + 18 – 16] = –3 square units
= 3 sq. units [numerically]Similarly, area of ACD
=1
2[4(2 – 0) + 5{0 – (–6)} + 4(–6 – 2)]
=1
2[8 + 30 – 32] = 3 sq. units
Clearly, area of ABD = area of ACDHence, a median of a triangle divides it into
two triangle of equal areas. Verified.
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OTHER IMPORTANT QUESTIONS
Q.1. The area of the triangle formed by(x, y + z), (y, z + x) and (z, x + y) is :
(a) x + y + z (b) xyz
(c) (x + y + z)2 (d) 0
Sol. (d) Required area
=1
2[x (z + x – x – y) + y (x + y – y – z)
+ z (y + z – z – x)
=1
2[xz – xy + xy – yz + zy – xz]
=1
2× 0 = 0.
Q.2. If points (x, 0), (0, y) and (1, 1) are
collinear, then1 1
+x y
is equal to :
(a) 1 (b) 2 (c) 0 (d) –1
Sol. (a) (x, 0), (0, y) and (1, 1) are collinear.
x (y – 1) + 0 (1 – 0) + 1 (0 – y) = 0
xy – x – y = 0
x + y = xy 1 1
+x y
= 1.
Q.3. If (x, 2), (–3, – 4) and (7, –5) arecollinear, then x is equal to :
(a) 60 (b) 63 (c) – 63 (d) – 60
Sol. (c) (x, 2), (–3, – 4) and (7, –5) are collinear.
x (– 4 + 5) – 3 (–5 – 2) + 7 (2 + 4) = 0
x + 21 + 42 = 0 x = –63
Q.4. If the area of the triangle formed by thepoints (a, 2a), (–2, 6) and (3, 1) is 5 square units,then a is equal to :
(a) 2 (b)3
5(c) 3 (d) 5
Sol. (a) We have, 5 =1
2[a (6 – 1) – 2 (1 – 2a)
+ 3 (2a – 6)]
10 = 5a – 2 + 4a + 6a – 18
15a = 30 a = 2
Q.5. If the area of a quadrilateral ABCD iszero, then the four points A, B, C and D are :
(a) collinear (b) not collinear
(c) nothing can be said
(d) none of these
Sol. (a) Area of quadrilateral ABCD = 0, means
A, B, C, D are collinear.
Q.6. The area of the triangle with verticesA(3, 0), B(7, 0) and C(8, 4) is :
(a) 14 (b) 28 (c) 8 (d) 6Sol. (c) Required area
=1
2[3(0 – 4) + 7(4 – 0) + 8(0 – 0)]
=1
2[–12 + 28] = 8 sq. units.
Q.7. If the points A(1, 2), O(0, 0) and C(a, b)are collinear, then : [2011 (T-II)]
(a) a = b (b) a = 2b(c) 2a = b (d) a = –b
Sol. (c) We have
1(0 – b) + 0(b – 2) + a(2 – 0) = 0
–b + 2a = 0 b = 2a
Q.8. Check whether the points (4, 5), (7, 6)and (6, 3) are collinear.
Sol. Area of the triangle formed by the points
(4, 5), (7, 6) and (6, 3)
=1
2[4 (6 – 3) + 7 (3 – 5) + 6 (5 – 6)]
=1
2[12 – 14 – 6] = – 4 0
Hence, the points are not collinear.
Q.9. Find the area of the triangle ABC withA(1, – 4), and the mid-points of sides through Abeing (2, –1) and (0, –1). [Imp.]
Sol. Let the coordinates of B and C be (x1, y1)
and (x2, y2) respectively.
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Then, 1 11+ –4 +,
2 2
x y
= (2, – 1) and 2 21+ –4 +,
2 2
x y
= (0, – 1)
1 + x1 = 4, – 4 + y1 = –2
and 1 + x2 = 0 and – 4 + y2 = –2
x1 = 3, y1 = 2 and x2 = –1, y2 = 2
Coordinates of B and C are (3, 2) and(–1, 2) respectively.
Area of ABC =1
2[1(2 – 2)
+ 3(2 + 4) – 1(– 4 – 2)]
=1
2[0 + 18 + 6] = 12 sq. units.
Q.10. Find the values of k if A(k + 1, 2k),B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Q.14. If three points (x1, y1), (x2, y2), (x3, y3)lie on the same line, prove that
2 3 3 1 1 2
2 3 3 1 1 2
– – –+ +
y y y y y y
x x x x x x = 0. [V. Imp.]
Sol. Points (x1, y1), (x2, y2) and (x3, y3) are
collinear. So, area of the triangle formed by thesepoints will be zero.
x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
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( ) ( )1 2 3 2 3 1
1 2 3 1 2 3
– –+
x y y x y y
x x x x x x+
( )3 1 2
1 2 3
–x y y
x x x= 0
[Dividing by x1, x2, x3]
2 3 3 1 1 2
2 3 1 3 1 2
– – –+ +
y y y y y y
x x x x x x= 0. Proved
Q.15. Four points A(6, 3), B(–3, 5), C(4, –2)
and D(x, 3x) are given in such a way that
( )( )
1=
2
ar ΔDBC
ar ΔABC, find x. [V. Imp.]
Sol. For DBC
x –3 4 x
3x 5 –2 3x
Area of DBC =1
2[(5x + 6 + 12x)
– (–9x + 20 – 2x)]
=1
2[(28x – 14)] = [14x – 7] = 7|2x – 1|
For ABC
6 –3 4 6
3 5 –2 3
Area of ABC
=1
2[(30 + 6 + 12) – (–9 + 20 – 12)]
=1
2[48 + 1] =
49
2.
Now,Area of DBC 1
=Area of ABC 2
∆∆
7 2 – 1 1 7
= 2 – 1 = ±49 2 4
2
xx⇒
2x =11
4or, 2x =
–3
4
x =11
8or x =
–3
8.
Q.16. If a b c, prove that the points(a, a2), (b, b2) and (c, c2) can never be collinear.
Sol. Let be the area of the triangle formed by
the points (a, a2), (b, b2), (c, c2);
a b c a
a2 b2 c2 a2
=1
2[(ab2 + bc2 + ca2) – (a2b + b2c + c2a)]
=1
2[(a2 c – a2 b) + (ab2 – ac2)
+ (bc2 – b2c)]
=1
2[–a2 (b – c) + a(b2 – c2) – bc (b – c)]
=1
2[{(b – c) (–a2 + a(b + c) – bc}]
=1
2[(b – c) (– a2 + ab + ac – bc)]
=1
2[(b – c) {–a(a – b) + c(a – b)}]
=1
2[(b – c) (a – b) (c – a)]
It is given that a b c 0.
Hence, given points are never collinear. Proved.
Q.17. The area of a triangle is 5 sq. units.Two of its vertices are (2, 1) and (3, –2). Thethird vertex lies on y = x + 3. Find the thirdvertex. [HOTS]
Sol. Let the third vertex be A(x, y). Other two
vertices of the triangle are B(2, 1) and C(3, –2).
x 2 3 x
y 1 –2 y
Area of ABC = 5 sq. units.
1
2|(x – 4 + 3y) – (2y + 3 – 2x)| = 5
1
2|(x – 4 + 3y – 2y – 3 + 2x)| = 5
1
2|3x + y – 7| = 5 3x + y – 7 = ± 10
3x + y – 17 = 0 or 3x + y + 3 = 0
It is given that the vertex A(x, y) lies on y = x + 3.
Solving 3x + y – 17 = 0 and y = x + 3, we get
x =7
2and y =
13
2
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Solving 3x + y + 3 = 0 and y = x + 3, we get
x =–3
2and y =
3
2Hence, the coordinates of the third vertex are
7 13 –3 3, or ,
2 2 2 2
.
Q.18. A(6, 1), B(8, 2) and C(9, 4) are threevertices of a parallelogram ABCD. If E is themid-point of DC, find the area of ADE.
Sol. Let the coordinates of D be (x, y). Then,
we know that the diagonals of a parallelogram bisecteach other.
mid-point of AC = mid-point of BD
15 5
,2 2
=
8 + 2 +,
2 2
x y
8 + 15
=2 2
xand
2 + 5=
2 2
y
x = 7 and y = 3
Coordinates of D are (7, 3).
Now, coordinates of E are7 + 9 3 + 4
,2 2
=7
8,2
Area of ADE
=1
2( )
7 76 3 – + 7 – 1
2 2
+ 8 1 – 3
sq. units.
=1
2
35–3 + – 16
2
sq. units
=1
2
3–
2
sq. units = –
3
4sq. units
=3
4sq. units.
Q.19. If the points A(1, –2), B(2, 3), C(a, 2)and D(– 4, –3) form a parallelogram, find thevalue of a and height of the parallelogram takingAB as base. [HOTS]
Sol. Diagonals of a parallelogram bisect each
other
mid-point of AC = mid-point of BD
1+ –2 + 2
,2 2
a =
2 – 4 3 – 3,
2 2
1 + a = –2 a = –3
Area of ADC =1
2[1 (– 3 – 2) – 4 (2 + 2)
+ a (– 2 + 3)] sq. units.
=1
2[– 5 – 16 + a] sq. units =
1
2[–21 – 3]
= 12 sq. unitsArea of the parallelogram = 2 × 12 sq. units
= 24 sq. units
AB = ( ) ( )2 21 – 2 + – 2 – 3
= 1 + 25 units = 26 units
Area of parallelogram = AB × height
24 = 26 × h
h =24 24 × 26 12 26
= =26 1326
units.
Q.20. If A(4, –6), B(3, –2) and C(5, 2) arethe vertices of ABC, then verify the fact that amedian of a triangle ABC divides it into twotriangles of equal areas. [2011 (T-II)]
Sol. Let D be the mid-point of BC. Then, the
coordinates of D are (4, 0).
We have
4 3 5 4
–6 –2 2 –6
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Area of ABC
=1
2|(4 × –2 + 3 × 2 + 5 × –6)
– (3 × –6 + 5 × –2 + 4 × 2)|
Area of ABC =1
2|(–8 + 6 – 30)
– (–18 – 10 + 8)|
Area of ABC =1
2|(–32 + 20)| = 6 sq. units
Also, we have
4 3 4 4
–6 –2 0 –6
Area of ABD
=1
2|{4 × (–2) + 3 × 0 + 4 × (–6)}
– {(3 × (–6) + 4 × (–2) + 4 × 0}|
Area of ABD =1
2|(–8 + 0 – 24)
– (–18 – 8 + 0)|
Area of ABD =1
2|(–32 + 26)| = 3 sq. units
Area of ABC 6 2
Area of ABD 3 1
∆= =
∆ Area of ABC = 2(Area of ABD)
PRACTICE EXERCISE 7.3A
Choose the correct option (Q.1 – 5) :
1. Area of the triangle with vertices (x, 0),(0, y) and (x, y) is :
(a) x + y (b) x – y (c)x
y(d)
2
xy
2. Area of the triangle with vertices (0, 0),(x, 0) and (x, y) is :
(a) xy (b) 2xy (c)2
xy(d)
3
xy
3. If (a, b), (0, a) and (a, 0) are collinearthen :
(a) a = 0 (b) b = 0
(c) either a = 0 or b = 0
(d) none of these
4. Area of the triangle having vertices(– 4, 2), (– 4, 6) and (– 4, –6) is :
(a) 7 sq. units (b) 10 sq. units
(c) 11 sq. units (d) 0
5. The area of the triangle whose vertices are(a, a + b), (b, a + b) and (a, b) is :
(a)2 +
2
a ab(b) a2 + ab
(c)2 –
2
a ab(d) a2 – ab
6. Find the area of the triangle whosevertices are :
(a) (4, 6), (2, 6), (0, 0)
(b) (–5, –3), (–6, –8), (0, 0)
(c) , , ,a a b b
b c a c
, (0, 0)
(d) (a, a + b), (b, a + b), (a, b)
7. The coordinates of A, B, C are (3, 4),(5, 2), (x, y) respectively. If area of ABC = 3,show that x + y = 10. [Imp.]
8. If the vertices of a ABC are A(1, m),B(4, –3) and C(–9, 7) and its area is 15 sq. units,find the value of m.
9. A and B are points (3, 4) and (5, –2). Findthe coordinates of the point P such that|PA| = |PB| and area of PAB = 10 sq. units.
10. Find the area of the quadrilateral whosevertices are :
(a) (1, 1), (7, –3), (12, 2), (7, 21)
(b) (0, 0), (4, 6), (2, 5), (5, 1)
11. Find the area of ABC, the coordinatesof the mid-points of whose sides are D(–1, –2),E(6, 1) and F(3, 5) respectively.
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12. If D1 5
– ,2 2
, E (7, 3) and F
7 7,
2 2
are
the mid-points of sides of ABC, find the area ofABC. [Imp.]
13. Find the value of x if the points (x, 8),(– 4, 2) and (5, –1) are collinear.
14. Show that the points (2, 4), (0, 1), (4, 7)are collinear.
15. Prove that the area of triangle whosevertices are (t, t – 2), (t + 2, t + 2) and(t + 3, t) is independent of t. [Imp.]
16. Prove that the points (a, b), (c, d) and(a – c, b – a) are collinear if ad = cb. [Imp.]
17. Find the relation between x and y if thepoints (x, y), (1, 2) and (7, 0) are collinear.
[2009]
18. Find the area of the triangle formed byjoining the mid-points of the sides of the trianglewhose vertices are (0, –1), (0, 1) are (0, 3).
[2009]
TEXTBOOK’S EXERCISE 7.4 (OPTIONAL)
Q.1. Determine the ratio in which the line2x + y – 4 = 0 divides the line segment joining thepoints A(2, –2) and B(3, 7). [2011 (T-II)]
Sol. Let the line 2x + y – 4 = 0 divide the line
segment joining the points A(2, –2) and B(3, 7) in theratio K : 1. Let the point of intersection be P.
Q.5. The Class X students of a secondaryschool in Krishinagar have been allotted arectangular plot of land for their gardeningactivity. Sapling of Gulmohar are planted on theboundary at a distance of 1 m from each other.There is a triangular grassy lawn in the plot asshown in figure. The students are to sow seeds offlowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates ofthe vertices of the triangle.
(ii) What will be the coordinates of thevertices of PQR if C is the origin ? Alsocalculate the areas of the triangles in thesecases. What do you observe?
Sol. (i) Taking A as origin, AD and AB as
coordinate axes, the coordinates of the vertices of thetriangle PQR are P(4, 6), Q(3, 2) and R(6, 5).
(ii) Taking C as origin, CB and CD as coordinateaxis, the coordinates of the vertices of triangles PQRare P(12, 2), Q(13, 6), R(10, 3).
Area of the triangle PQR taking A as origin
=1
2[x1 (y2 – y1) + x2 (y3 – y1) + x3 (y1 – y2)]
=1
2[4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
=1
2[–12 – 3 + 24]
=9
2sq. units = 4.5 sq. units … (i)
Area of PQR taking C as origin
=1
2[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
=1
2[36 + 13 – 40]
=9
2sq. units = 4.5 sq. units … (ii)
From (i) and (ii), we find that areas are same.
Q.6. The vertices of a ABC are A(4, 6),B(1, 5) and C(7, 2). A line is drawn to intersectsides AB and AC at D and E respectively, such
thatAD AE 1
= =AB AC 4
. Calculate the area of the
ADE and compare it with the area of ABC.[HOTS]
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Sol. We have,
AD AE 1= =
AB AC 4
AB AC 4
= =AD AE 1
AD + DB AE + EC
=AD AE
= 4
1 +DB
AD= 1 +
EC
AE= 4
DB
AD=
EC
AE= 3
AD AE 1= =
DB EC 3
D and E divide AB and AC respectively in theratio 1 : 3.
So, the coordinates of D and E are
1+12 5 +18 13 23, = ,
1+ 3 1+ 3 4 4
and7 +12 2 +18 19
, = , 51+ 3 1+ 3 4
respectively..
For ADE :
413
4
19
44
623
45 6
Area of ADE
=1
2
23 13 194× + × 5 + × 6
4 4 4
–13 19 23
× 6 + × + 4 × 54 4 4
=1
2
92 65 114+ +
4 4 4
78 437– + + 20
4 16
=1
2
271 1069–
4 16
=
1
2×
15
16
=15
32sq. units.
Also, for ABC :
4 1 7 4
6 5 2 6
Area of ABC =1
2[(4 × 5 + 1 × 2 + 7 × 6)
– (1 × 6 + 7 × 5 + 4 × 2)]
=1
2[(20 + 2 + 42) – (6 + 35 + 8)]
=1
2[64 – 49] =
15
2sq. units.
15Area of ADE 132= =
15Area of ABC 16
2
∆∆
.
Hence, Area of ADE : Area of ABC = 1 : 16.
Q.7. Let A(4, 2), B(6, 5) and C(1, 4) be thevertices of ABC.
(i) The median from A meets BC at D. Findthe coordinates of the point D.
(ii) Find the coordinates of the point P onAD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and Ron medians BE and CF respectively such thatBQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are thevertices of ABC, find the coordinates of thecentroid of the triangle.
Sol. Given vertices of ABC are A(4, 2), B(6, 5)
and C(1, 4).
(i) AD is the median from vertex A, D is the mid-point of BC.
Using mid-point formula the coordinates of D are
6 + 1 5 + 4 7 9, i.e., ,
2 2 2 2
.
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(ii) Let P(x, y) be the point on AD such thatAP : PD = 2 : 1
Then,
72 +1× 4
7 + 4 112= = =
2 +1 3 3x
92 +1× 2
9 + 2 112= = =
2 +1 3 3y
Hence, coordinates of point P are11 11
,3 3
(iii) Since, BE and CF are medians, E and F aremid-points of AC and AB respectively.
So, x1 =4 +1 5
=2 2
, y1 =4 + 2 6
=2 2
= 3
Coordinates of point E are5
, 32
.
Similarly, x2 =4 + 6 10
=2 2
= 5,
y2=5 + 2 7
=2 2
Coordinates of point F are7
5,2
Q divides BE such that BQ : QE = 2 : 1
Coordinates of Q are
52× + 6 × 1
2×3 + 1 × 52 ,2 +1 2 +1
=5 + 6 6 + 5
,3 3
=
11 11,
3 3
Similarly, R divides CF such that CR : RF= 2 : 1
Coordinates of R are 11 11,
3 3
.
(iv) From above, it is clear that coordinates of P,Q and R are same. This point is known as centroid oftriangle, which divides each median in the ratio 2 : 1.
(v) The vertices of ABC are A(x1, y1), B(x2, y2)and C(x3, y3). D is the mid-point of BC.
So, coordinates of D are 2 3 2 3+ +, .
2 2
x x y y
Let G(x, y) be the centroid of ABC.
Then G divides the median AD in the ratio 2 : 1
So, x =
2 31
+2 × +
22 +1
x xx
and
y =
2 31
+2 × +
22 +1
y yy
x =1 2 3+ +
3
x x xand
y=1 2 3+ +
3
y y y
Hence, coordinates of the centroid of ABC
are 1 2 3 1 2 3+ + + +,
3 3
x x x y y y.
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Q.8. ABCD is a rectangle formed by thepoints A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1).P, Q, R and S are the mid-points of AB, BC, CDand DA respectively. Is the quadrilateral PQRSa square? a rectangle? or a rhombus? Justifyyour answer. [2011 (T-II)]
Sol. The vertices of the given rectangle are
A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R andS are the mid-points of side AB, BC, CD and DArespectively.
Coordinates of P are
–1 –1 –1 + 4 3, = –1,
2 2 2
Coordinates of Q are–1 + 5 4 + 4
,2 2
= (2, 4)
Coordinates of point R are 5 + 5 –1 + 4,
2 2
=3
5,2
Coordinates of point S are–1 + 5 –1 – 1
,2 2
= (2, –1)
PQ = ( )2
2 32 + 1 + 4 –
2
=
25 619 + =
4 2
QR = ( )2
2 35 – 2 + – 4
2
=25 61 61
9 + = =4 4 2
RS = ( )2
2 3 252 – 5 + –1– = 9 +
2 4
=61 61
=4 2
SP = ( )2
2 3 252 + 1 + –1 – = 9 +
2 4
=61 61
=4 2
PR = ( )2
2 3 35 + 1 + – = 6
2 2
QS = ( ) ( )2 22 – 2 + 4 + 1 = 5
Hence, we see that
PQ = QR = RS = SP
All the sides are equal.
But, PR QS (diagonals are not equal)
Hence, PQRS is a rhombus.
B. FORMATIVE ASSESSMENT
Activity-1
Objective : To verify the distance formula and section formula. Or to verify the following :
(i) The distance between the points (x1, y1) and (x2, y2) is ( ) ( )2 22 1 2 1– + –x x y y
(ii) The coordinates of the point P, which divides the line segment joining the points
A(x1, y1) and B(x2, y2) in the ratio m : n are2 1 2 1+ +
,+ +
mx nx my ny
m n m n
.
(iii) The coordinates of the mid-points P of the line segment joining the points A (x1, y1) and
B(x2, y2) are 1 2 1 2+ +,
2 2
x x y y
.
Materials Required : Graph paper/squared paper, geometry box, etc.Procedure : 1. Take a 1 cm squared paper and on it draw the coordinate axes XOX' and YOY'.
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2. Plot the points A (–3, 1), B (5, 1) and C (1, 8) on the squared paper and join AB, BC and ACto get a ABC.
3. Now using a pair of compasses, find the mid-points of AC and BC. Mark these mid-points asP and Q respectively.
4. Mark the point R on AB such that AR =1
4AB or R divides AB in the ratio 1 : 3. Similarly,,
mark S on AB such that AS =3
4AB or S divides AB in the ratio 3 : 1.
5. From the graph paper write the coordinates of P, Q, R and S.
Coordinates of
P (–1, 9/2)Q (3, 9/2)R (–1, 1)S (3, 1)
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Observations :
1. Using a ruler, measure the length of AB, BC and CA
We have AB = BC = CA = 8 cm.
2. Now, using formula
AB = ( ) ( )2 25 + 3 + 1 – 1 cm = 8 cm
BC = ( ) ( )2 21 – 5 + 8 – 1 cm = 16 + 49 cm = 65 cm = 8.06 cm = 8 cm
CA = ( ) ( )2 21+ 3 + 8 –1 cm = 16 + 49 cm = 65 cm = 8.06 cm = 8 cm
We see that in both the cases, the length of each side comes out to be 8 cm.
3. Since P and Q are mid-points of AC and BC respectively, therefore, using formula, the
coordinates of P are1 – 3 8 +1 9
, or –1,2 2 2
the coordinate of Q are
1+ 5 8 +1 9, or 3,
2 2 2
Also, from the table : the coordinates of P and Q are same as obtained above.
4. Since R divides AB in the ratio 1 : 3, therefore, coordinates of R (using formula) are
( )1×5 + 3× –3 1×1+ 3×1,
1+ 3 1+ 3
or (–1, 1).
Similarly, S divides AB in the ratio 3 : 1, therefore, coordinates of S (using formula) are :
( )3×5 +1× –3 3×1+1×1,
3 +1 3 +1
or (3, 1). Also, from the table, the coordinates of R and S are same
as obtained above.
Conclusion : From the above activity it is verified that :
(i) the distance of the line segments joining the points (x1, y1) and (x2, y2) is given by
( ) ( )2 22 1 2 1– + –x x y y
(ii) If a point P divides the line segment joining the points (x1, y1) and (x2, y2) in the ratio m :
n, then the coordinates of P are2 1 2 1+ +
,+ +
mx nx my ny
m n m n
.
(iii) The coordinates of the mid-point of the line segment joining the points (x1, y1) and (x2, y2)
are1 2 1 2+ +
,2 2
x x y y .
Do Yourself : Draw three different triangles on a paper and in each case verify the above formulae.
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Activity-2
Objective :To verify the following formula :The area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is given by1
2[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
Materials Required : Graph paper/Squared paper, geometry box etc.
Procedure : 1. Take a 1 cm squared paper and on it draw the coordinate axes XOX' and YOY'.
Case I.1. Plot the points A (–3, 0), B (4, 0) and C(0, 4) on the squared paper and join AB, BC and AC
to get a scalene triangle ABC.2. Find AB, BC and AC using the distance formula.
AB = ( ) ( )2 2– 3 – 4 + 0 – 0 cm
= 49 cm = 7 cm
BC = ( ) ( )2 24 – 0 + 0 – 4 cm
= 16 + 16 cm = 4 2 cm
AC = ( ) ( )2 2– 3 – 0 + 0 – 4 cm
= 9 + 16 cm = 5 cm
3. Find the area of ABC usingHeron's formula
Here, a = 7, b = 4 2 , c = 5
s =7 + 4 2 + 5
= 6 + 2 22
Area of ABC = – – –s s a s b s c
6 + 2 2 6 + 2 2 – 7 6 + 2 2 – 4 2 6 + 2 2 – 5 cm2
= 6 + 2 2 2 2 – 1 6 – 2 2 2 2 + 1 cm2
= 6 + 2 2 6 – 2 2 2 2 – 1 2 2 + 1 cm2
= ( )( ) 236 – 8 8 – 1 cm = 2 228× 7 cm = 14 cm
4. Now, find the area of the triangle ABC, using the formula
1
2[x1(y2 – y3) – x2(y3 – y1) + x3(y1 – y2)]
Area of ABC =1
2[–3(0 – 4) + 4(4 – 0) + 0(0 – 0)] cm2 =
1
2[12 + 16] cm2 = 14 cm2.
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Observations : From (3) and (4), we see that the area of ABC comes out to be same on boththe cases.
Case II.
1. Plot the points P (–2, 3), Q (–2, 0) and R (2, 0) and join PQ, QR and PR to get the right trianglePQR.
Observations : From (3) and (4) above we see that area of PQR comes out to be same in boththe cases.
Conclusion : From the above activity it is verified that the area of a triangle having vertices(x1, y1), (x2, y2) and (x3, y3) is given by
1
2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Investigation
On a squared / graph paper, draw a rectangle of dimensions 4 × 3. Draw one of the diagonalsof the rectangle. The diagonal passes through 6 squares.
Now on a squared paper draw several rectangles of different sizes. The length (l) and breadth (b)of each rectangle must be a whole number of squares with a common factor of 1 only. For examplesides 4 squares by 9 squares is acceptable but 4 squares by 6 squares is not because 4 and 6 havea common factor 2.
For each rectangle draw a diagonal and count the number of squares through which the diagonal passes.
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The dimensions of a rectangle is 120 × 91. How manysquares will the diagonal pass through?
If the dimensions of a rectangle are m × n, where mand n are co-primes, then how many squares will thediagonal pass through?
Puzzle
A merchant has nine gold coins which look identical butin fact one of the coins is an underweight fake. Investigatehow the merchant can use only a balance to find the fakecoin in just two weighings.
If m and n are the dimensions of the rectangle, where m and n are co-prime, then the diagonal will passthrough m + n – 1 squares.
PuzzlePlace three coins on one pan and three coins on another pan of a balance. If the pans balance each other,
it means the under weight coin is among the remaining three coins. Now, remove these coins and place one coinof the remaining three coins on one pan and other on the other pan of the balance. If the pans balance each other,it means the remaining third coin is underweight. While weighing six coins, above, if the pans do not balanceeach other, then you can clearly identify the pan in which the underweight coin is placed. Then proceed further.
