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1. Cartesian Coordinate system and Quadrants
2. Distance formula3. Area of a triangle4. Collinearity of three points5. Section formula6. Special points in a triangle7. Locus and equation to a locus8. Translation of axes - shift of
origin9. Translation of axes - rotation of
axes
Session Objectives
Coordinates
XX’
Y
Y’
O
Origin
1 2 3 4
+ve direction
-1-2-3-4
-ve direction
-1
-2
-3 -ve d
irect
ion
1
2
3
+ve d
irect
ion
X-axis : X’OX
Y-axis : Y’OY
Quadrants
XX’ O
Y
Y’
III
III IV
(+,+)(-,+)
(-,-) (+,-)Q : (1,0) lies in which Quadrant?
Ist? IInd?
A : None. Points which lie on the axes do not lie in any quadrant.
Distance Formula
P(x 1, y 1
)
Q(x 2, y 2
)
x1
XX’
Y’
O
Y
x2
y1
y2
N PQN is a right angled .
PQ2 = PN2 + QN2
2 22 1 2 1PQ x x y y
y2-y
1(x2-x1)
PQ2 = (x2-x1)2+(y2-y1)2
Area of a Triangle
XX’
Y’
O
Y A(x1, y1)
C(x3, y3)
B(x
2,
y2)
M L N
Area of ABC =
Area of trapezium ABML + Area of trapezium ALNC- Area of trapezium BMNC
Area of a Triangle
XX’Y’O
YA(x1, y1)
C(x3, y3)
B(x
2,
y2)
M L N
Area of trapezium ABML + Area of trapezium ALNC- Area of trapezium BMNC
1 1 1BM AL ML AL CN LN BM CN MN
2 2 2
2 1 1 2 1 3 3 1 2 3 3 21 1 1
y y x x y y x x y y x x2 2 2
1 1
2 2
3 3
x y 11
x y 12
x y 1
Sign of Area : Points anticlockwise +ve
Points clockwise -ve
Area of Polygons
Area of polygon with points Ai (xi, yi)
where i = 1 to n
2 21 1 n 1 n 1 n n
3 32 2 n n 1 1
x yx y x y x y1. . .
x yx y x y x y2
Can be used to calculate area of Quadrilateral,
Pentagon, Hexagon etc.
Collinearity of Three Points
Method II :
Use Area of Triangle
A (x1, y1)
B (x2, y2)
C (x3, y3)
Show that1 1
2 2
3 3
x y 1
x y 1 0
x y 1
Section Formula – Internal Division
A(x 1, y 1
)
B(x 2, y 2
)
XX’
Y’
O
Y
P(x, y)
m
n:
L N M
H
K
Clearly AHP ~ PKBAP AH PHBP PK BK
1 1
2 2
x x y ymn x x y y
2 1 2 1mx nx my nyP ,
m n m n
Section Formula – External Division
XX’
Y’
O
Y
A(x 1, y 1
)
P(x, y)
B(x 2, y 2
)
L N M
H
K
Clearly PAH ~ PBKAP AH PHBP BK PK
1 1
2 2
x x y ymn x x y y
2 1 2 1mx nx my nyP ,
m n m n
P divides AB externally in ratio m:n
Centroid
Intersection of medians of a triangle is called the centroid.
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
Centroid is always denoted by G.
Centroid
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1
2 3 2 31 1
x x y yx 2 y 2
2 2L ,1 2 1 2
Centroid
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1
1 3 1 32 2
x x y yx 2 y 2
2 2M ,1 2 1 2
Centroid
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1
1 2 1 23 3
x x y yx 2 y 2
2 2N ,1 2 1 2
Centroid
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
1 2 3 1 2 3x x x y y yL ,
3 3
1 2 3 1 2 3x x x y y yM ,
3 3
1 2 3 1 2 3x x x y y yN ,
3 3
We see that L M N G
Medians are concurrent at the centroid, centroid divides medians in
ratio 2:1
Centroid
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF G
2 3 2 3x x y yD ,
2 2
1 3 1 3x x y yE ,
2 2
1 2 1 2x x y yF ,
2 2
1 2 3 1 2 3x x x y y yL ,
3 3
1 2 3 1 2 3x x x y y yM ,
3 3
1 2 3 1 2 3x x x y y yN ,
3 3
We see that L M N G
Centroid
1 2 3 1 2 3x x x y y yG ,
3 3
Incentre
Intersection of angle bisectors of a triangle is called the incentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF I
Incentre is the centre of the incircle
Let BC = a, AC = b, AB = c
AD, BE and CF are the angle bisectors of A, B and C respectively.
BD AB bDC AC c
2 3 2 3bx cx by cy
D ,b c b c
Incentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF I
BD AB bDC AC c
2 3 2 3bx cx by cy
D ,b c b c
AI AB AC AB AC c bNow,
ID BD DC BD DC a
2 3 2 31 1
bx cx by cyax b c ay b c
b c b cI ,a b c a b c
1 2 3ax bx cxI
a b c
Similarly I can be derived using E and F also
Incentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF I
BD AB bDC AC c
2 3 2 3bx cx by cy
D ,b c b c
AI AB AC AB AC c bNow,
ID BD DC BD DC a
2 3 2 31 1
bx cx by cyax b c ay b c
b c b cI ,a b c a b c
1 2 3ax bx cxI
a b c
Angle bisectors are concurrent at the incentre
Excentre
Intersection of external angle bisectors of a triangle is called the excentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF
E
Excentre is the centre of the excircleEA = Excentre opposite A
1 2 3 1 2 3A
ax bx cx ay by cyE ,
a b c a b c
Excentre
Intersection of external angle bisectors of a triangle is called the excentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF
E
Excentre is the centre of the excircleEB = Excentre opposite B
1 2 3 1 2 3B
ax bx cx ay by cyE ,
a b c a b c
Excentre
Intersection of external angle bisectors of a triangle is called the excentre
A(x1, y1)
B(x2, y2) C(x3, y3)D
EF
E
Excentre is the centre of the excircleEC = Excentre opposite C
1 2 3 1 2 3C
ax bx cx ay by cyE ,
a b c a b c
Cirumcentre
Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre.
OA = OB = OC
= circumradius
A
B
CO
The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.
Orthocentre
Intersection of altitudes of a triangle is called the orthocentre.
A
B C
H
Orthocentre is always denoted by H
We will learn to find coordinates of Orthocentre after we learn straight lines
and their equations
Cirumcentre, Centroid and Orthocentre
The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear.
O
H
G
1 : 2
G divides OH in the ratio 1:2
Locus – a Definition
The curve described by a point which moves under a given condition or conditions is called its locus
e.g. locus of a point having a constant distance from a fixed point :
Circle!!
Locus – a Definition
The curve described by a point which moves under a given condition or conditions is called its locus
e.g. locus of a point equidistant from two fixed points :
Perpendicular bisector!!
Equation to a Locus
The equation to the locus of a point is that relation which is satisfied by the coordinates of every point on the locus of that point
Important :A Locus is NOT an equation. But it is associated with an
equation
Equation to a Locus
Algorithm to find the equation to a locus :
Step I : Assume the coordinates of the point whose locus is to be found to be (h,k)
Step II : Write the given conditions in mathematical form using h, k
Step III : Eliminate the variables, if any
Step IV : Replace h by x and k by y in Step III. The equation thus obtained is the required equation to locus
Illustrative Example
Find the equation to the locus of the point equidistant from A(1, 3) and B(-2, 1)
Let the point be P(h,k)
PA = PB (given)
PA2 = PB2
(h-1)2+(k-3)2 = (h+2)2+(k-1)2
6h+4k = 5
equation of locus of (h,k) is 6x+4y = 5
Solution :
Illustrative Example
A rod of length l slides with its ends on perpendicular lines. Find the locus of its midpoint.
Let the point be P(h,k)
Let the lines be the axes
Let the rod meet the axes at
A(a,0) and B(0,b)
h = a/2, k = b/2
Also, a2+b2 = l2
4h2+4k2 = l2
equation of locus of (h,k) is 4x2+4y2 = l2
B(0,b)
A(a,0)O
P(h,k)
Solution :
Shift of Origin
XX’
Y’
O
Y
O’(h,k)
P(x,y)
x
yX Y
Consider a point P(x, y)
Let the origin be shifted to O’ with coordinates (h, k) relative to old axes
Let new P (X, Y)
x = X + h, y = Y + k
X = x - h, Y = y - kO (-h, -k) with reference to new axes
Illustrative Problem
Show that the distance between two points is invariant under translation of the axes
Let the points have vertices
A(x1, y1), B(x2, y2)
Let the origin be shifted to (h, k)
new coordinates : A(x1-h, y1-k), B(x2-h, y2-k)
2 21 2 1 2Old dist. (x x ) (y y )
2 21 2 1 2& New dist. (x h x h) (y h y h)
= Old dist.
Solution :
Rotation of Axes
XX’
Y’
O
Y
P(x,y)
x
yConsider a point P(x, y)
Let the axes be rotated through an angle .Let new P (X, Y) make an angle with the new x-axis
X
X’
Y’
O
Y
X
X’Y’
O
Y
X
X’ Y’
O
Y
X
X’Y’
O
Y
Y
X
xcos ,
R
R
ysin ,
R
Ysin ,
R
Xcos
R
Rotation of Axes
xcos ,
R y
sin ,R
Y
sin ,R
X
cosR
xcos cos sin sin
R
ysin cos cos sin
R
X Y xcos sin
R R R
X Y ysin cos
R R R
x X cos Y sin
y X sin Y cos
X x cos y sin
Y y cos x sin
Class Exercise - 1
If the segments joining the pointsA(a,b) and B(c,d) subtend an angle at the origin, prove that
2 2 2 2
ac bdcos
a b c d
Solution
On simplifying, 2 2 2 2
ac bdcos
a b c d
Let O be the origin.
OA2 = a2+b2, OB2 = c2+d2, AB2 = (c-a)2+(d-b)2
Using Cosine formula in OAB, we have
AB2 = OA2+OB2-2OA.OBcos
2 2 2 2 2 2 2 2 2 2c a d b a b c d 2 a b c d cos
Class Exercise - 2
Four points A(6,3), B(-3,5), C(4,-2)and D(x,3x) are given such that
Find x.DBC 1ABC 2
Given that ABC = 2DBC
6 3 1 x 3x 1
3 5 1 2 3 5 1
4 2 1 4 2 1
6 5 2 3 4 3 1 6 20 2 x 5 2 3x 4 3 1 6 20
2 28x 14 49 49
28x 142
11 3x or x
8 8
Solution :
Class Exercise - 3
If a b c, prove that (a,a2), (b,b2) and (c,c2) can never be collinear.
Let, if possible, the three points be collinear.
2
2
2
a a 11
b b 1 02
c c 1
R2 R2-R1, R3 R3- R2
2
2 2
2 2
a a 1
b a b a 0 0
c b c b 0
2a a 1
b a c b 1 b a 0 0
1 c b 0
Solution :
Solution Cont.R2 R2-R3
2a a 1
b a c b 0 a c 0 0
1 c b 0
b a c b c a 0
This is possible only if a = b or b = c or c = a.
But a b c. Thus the points can never be collinear.
Q.E.D.
Class Exercise - 4
Three vertices of a parallelogram taken in order are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b). Find the fourth vertex.
Let the fourth vertex be (x,y).
Diagonals bisect each other.
a b a b 2a b x a b a b 2a b yand
2 2 2 2
the required vertex is (-b,b)
Solution :
Class Exercise - 5
If G be the centroid of ABC and P be any point in the plane, prove that PA2+PB2+PC2=GA2+GB2+GC2+3GP2.
Let A (x1,y1), B (x2,y2), C (x3,y3), P (p,0)
LHS = (x1-p)2+y12+(x2-p)2+y2
2+(x3-p)2+y32
= (x12+y1
2)+(x22+y2
2)+(x32+y3
2)+3p2-2p(x1+x2+x3)
=GA2+GB2+GC2+3GP2
=RHS
Choose a coordinate system such that G is the origin and P lies along the X-axis.
Q.E.D.
Solution :
Class Exercise - 6
The locus of the midpoint of the portion intercepted between the axes by the line xcos+ysin = p, where p is a constant, is
2 2 22 2 2
2 22 2 2 2
1 1 4(a) x y 4p (b)
x y p
4 1 1 2(c) x y (d)
p x y p
Solution
Let the line intercept at the axes at A and B. Let R(h,k) be the midpoint of AB.
p pR h,k ,
2cos 2sin
p psin , cos
2k 2h
2 2
2 2
p p1
4k 4h 2 2 2
1 1 4Locus
x y p
Ans : (b)
Class Exercise - 7
A point moves so that the ratio of its distance from (-a,0) to (a,0) is 2:3. Find the equation of its locus.
Let the point be P(h,k). Given that
2 2
2 2
h a k 23h a k
2 2
2 2
h a k 49h a k
2 2 2
2 2 2
h 2ah a k 49h 2ah a k
2 2 25h 26ah 5k 5a 0
2 2 2
the required locus is
5x 26ax 5y 5a 0
Solution :
Class Exercise - 8
Find the locus of the point such that the line segments having end points (2,0) and (-2,0) subtend a right angle at that point.
Let A (2,0), B (-2,0)
Let the point be P(h,k). Given that2 2 2PA PB AB 2 2 22 2h 2 k h 2 k 2 2
2 22h 2k 8 16
2 2
the required locus is
x y 4
Solution :
Class Exercise - 9
Find the coordinates of a point where the origin should be shifted so that the equation x2+y2-6x+8y-9 = 0 will not contain terms in x and y. Find the transformed equation.
Let the origin be shifted to (h,k). The given equation becomes (X+h)2+(Y+k)2-6(X+h)+8(Y+k)-9 = 0
Or, X2+Y2+(2h-6)X+(2k+8)Y+(h2+k2-6h+8k-9) = 0
2h-6 = 0; 2k+8 = 0 h = 3, k = -4.
Thus the origin is shifted to (3,-4).
Transformed equation is X2+Y2+(9+16-18-32-9) = 0
Or, X2+Y2 = 34
Solution :
Class Exercise - 10
Through what angle should the axes be rotated so that the equation 11x2+4xy+14y2 = 5 will not have terms in xy?
Let the axes be rotated through an angle . Thus equation becomes
2
2
11 X cos Y sin 4 X cos Y sin X sin Y cos
14 X sin Y cos 5
Solution :
Solution Cont.
Therefore, the required angle is
cos 2sin 2cos sin 0
1tan or tan 2
2
1 11tan or tan 2
2
2 2
2 2
2 2
Or, 11cos 4sin cos 14sin X
4cos 6sin cos 4sin XY
11sin 4sin cos 14cos 5
2 22cos 3sin cos 2sin 0