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COORDINATE GEOMETRY

1. Distance Formula.
Distance between points A(x1, y1) and B(x2, y2 ) is given by
AB = (x2 − x1)2 + (y2 − y1)2 .
• Distance of a point P(x, y ) from origin O(0, 0) is OP = x 2 + y 2 .

For example:
Find the distance between the points P(–6, 7) and Q(–1, –5)
Here, x1 = –6, y1 = 7 and x2 = –1, y2 = –5
 PQ = (x2 − x1)2 + (y2 − y1)2
 PQ = (−1+ 6)2 + (−5 − 7)2

 PQ = 25 + 144 = 169 = 13

2. Three points will from:
(i) A right angle triangle, if sum of the squares of any two sides is equal to square of
third side.
(ii) An equilateral triangle, if all the three sides are equal.
(iii) An isosceles triangle, if any two sides are equal.
(iv) A collinear or a line, if sum of two sides is equal to third side.

For example:
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right-
angled triangle.
Let A(–3, 0), B(1, –3) and C(4, 1) be the given points. Then,

AB = [1− (−3)]2 + (−3 − 0)2

= 42 + (−3)2 = 16 + 9
= 5 units
BC = (4 − 1)2 + (1+ 3)2

= 9 + 16 = 5 units
and,CA = (4 + 3)2 + (1− 0)2

= 49 + 1 = 5 2 units
 AB = BC.
Therefore, ABC is isosceles.

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Also, AB2 + BC2 = 25 + 25 = (5 2)2 = CA2
ABC is right-angled at B. Thus, ABC is a right-angled isosceles triangle.

3. Four points will form:
(i) A rhombus, if all the four sides are equal.
(ii) A square, if all the four sides and diagonals are equal.
(iii) A parallelogram, if opposite sides are equal and diagonals bisect each other.
(iv) A rectangle, if opposite sides and diagonals are equal.

For example:
Show that four points (0, –1), (8, 3), (6, 7) and (−2, 3) are the vertices of a rectangle.
Let A(0 –1), B(6, 7), C(–2, 3) and D(8, 3) be the given points. Then,

AB = (8 − 0)2 + (3 + 1)2 = 64 + 16 = 4 5 D(−2, 3) C(6, 7)
DC = (6 + 2)2 + (7 − 3)2 = 64 + 16 = 4 5

BC = (8 − 6)2 + (3 − 7)2 = 6 + 16 = 2 5 A(0, –1) B(8, 3)
and, AD = (−2 − 0)2 + (3 + 1)2 = 4 + 16 = 2 5

Now, AC = (6 − 0)2 + (7 + 1)2 = 36 + 64 = 10

and, BD = (8 + 2)2 + (3 − 3)2 = 100 + 0 = 10

AC 2 = AB 2 + BC 2 = AD2 + DC 2
100 = 80 + 20 = 20 + 80
BD2 = BA2 + AD2 = BC 2 + CD 2
100 = 80 + 20 = 20 + 80
Hence, ABCD is a rectangle.

4. Section formula:
(i) Internal division
A(x1, y1) and B(x2, y2 ) are two points and P(x, y ) divides AB internally in the
ratio m : n, then

A m •n B
(x1, y1) P(x, y) (x2, y2)

x = mx 2 + nx 1 , y = my 2 + ny 1
m+n m+n

For example:

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Find the coordinates of the point which divides the line segment joining the points (6,
3) and (–4, 5) in the ratio 3 : 2 internally.

Let P(x, y) be the required point. Then,

x = 3  −4 + 2  6 and y = 3  5 + 2  3  x = 0 and y = 21
3+2 3+2 5

32

A(6, 3) P(x, y) B(–4, 5)

So, the coordinates of P are (0, 21/5).

(ii) Mid point formula
Coordinates of mid-point of AB are

x = x1 + x2 , y = y1 + y2 .
22

For example:

The coordinates of one end point of a diameter AB of a circle are A(4, –1) and the
coordinates of the centre of the circle are C(1, –3). Find the coordinates of B.

In the figure A(4, –1) and B(a, b) are the end A C B
points of the given diameter AB, and C(1, –3) is (4, –1) (1, –3) (a, b)
the centre of the circle. Then, C is the midpoint of
AB.

By the midpoint formula, the coordinates of C are

 4 + a , − 1+ b  .
2 2 

But, the coordinates of C are given as (1, –3).

 4 + a = 1and − 1+ b = −3
22

 4 + a = 2 and –1 + b = –6

 a = –2 and b = –5
Hence, the required point is B(–2, –5).

(iii) If we have to find the ratio in which the join of two points is divided according to
some given condition, then we take ratio as k : 1 as  kx2 + x1 , ky 2 + y1  .
 k +1 k +1 

For example:

In what ratio does the point C(3/5, 11/5) divide the line segment joining the points A(3,
5) and B(–3, –2)?.

Let the point C divide AB in the ratio k : 1. Then the coordinates of C are

 − 3k + 3 , − 2k + 5  k 1 B(–3, –2)
 k +1 k +1  A(3, 5) C

But, the coordinates of C are given as (3/5, 11/5).

 − 3k + 3 = 3 and − 2k + 5 = 11
k +1 5 k +1 5

 – 15k + 15 = 3k + 3 and – 10k + 25 = 11k + 11

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 18k = 12 and 21k = 14
 k = 2 . Hence, the point C divides AB in the ratio 2 : 3.

3

5. Area of triangle:

Area of ABC , whose vertices are A(x1, y1) , B(x2, y 2 ) and C(x3, y 3 ) is

= 1 x1(y 2 − y 3 ) + x2 (y 3 − y1) + x3(y1 − y2 )
2

For example:
Find the area of a triangle whose vertices are (1, –1), (–4, 6) and (–3, –5).
The area of the triangle formed by the vertices A(1, –1), B(–4, 6) and C(–3, –5) is given
by

1 [1(6 + 5) + (−4)(−5 + 1) + (−3)(−1− 6)] = 1 (11 + 16 + 21) = 24
22

So, the area of the triangle is 24 square units.

6. Condition for collinear points
Three points A(x1, y1), B(x2, y 2 ) and C(x3, y 3 ) are collinear or lie on a line if any one
of the following holds:
(i) AB + BC = AC
(ii) AC + CB = AB
(iii) CA + AB = CB

For example:
Show that the points (1, –1), (5, 2) and (9, 5) are collinear.
Let A(1, – 1), B(5, 2) and (9, 5) be the given points. Then, we have

AB = (5 − 1)2 + (2 + 1)2 = 16 + 9 = 5

BC = (5 − 9)2 + (2 − 5)2 = 16 + 9 = 5

and AC = (1− 9)2 + (−1− 5)2 = 64 + 36 = 10

Clearly, AC = AB + BC

Hence, A, B, C are collinear points.

(iv) If we have to show that three given points are collinear using area of a triangle,
then the area of triangle formed by them is 0.

1 | x1(y 2 − y3 ) + x2(y3 − y1) + x3 (y1 − y 2 ) |= 0
2

For example:
Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
Since the given points are collinear, the area of the triangle will be 0,

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 1 [2(k + 3) + 4(−3 − 3) + 6(3 − k )] = 0
2

 1 [2k + 6 − 24 + 18 − 6k] = 0
2

 1 (−4k ) = 0
2

Therefore, k = 0

• If we have to find area of a quadrilateral, then we divide the quadrilateral into
appropriate number of triangles and after finding areas of triangles, add them.

For example:

If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the
area of the quadrilateral ABCD.

By joining B to D, two triangles ABD and BCD are D(4, 5) C(–1, –6)
formed.

Now the area of ABD

= 1 [−5(−5 − 5) + (−4)(5 − 7) + 4(7 + 5)] A(–5,7) B(–4, –5)
2

= 1 (50 + 8 + 48) = 106 = 53 square units
22

Also, the area of BCD

= 1 [−4(−6 − 5) − 1(5 + 5) + 4(−5 + 6)]
2

= 1 (44 − 10 + 4) = 19 square units
2

So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.

• If in question, it is given that the area of a triangle is A square units and we have
to find some condition, then we must take two cases as  A after removing the
modulus sign from the formula of area of triangle.

For example:

Find the value of k for which the area formed by the triangle with vertices

A(k,2k ),B(− 2,6) and C(3,1) is 5 square units.

The vertices of the given ABC are A(k,2k ),B(− 2,6) and C(3,1) .

 x1 = k, y1 = 2k, x2 = −2, y 2 = 6 and x3 = 3, y3 = 1

Area of ABC = 1 x1(y 2 − y3 )+ x2 (y 3 − y1)+ x3 (y1 − y 2 )
2

= 1 k(6 − 1) − 2(1− 2k ) + 3(2k − 6)

2

= 1 5k − 2 + 4k + 6k − 18
2

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= 1 15k − 20 sq units.
2

But, area of ABC = 5 sq units [given]

 1 15k − 20 = 5  15k − 20 = 10
2

 (15k − 20) = 10 or (15k − 20) = −10

Hence,  k = 2 or k = 2 .
3

k = 2 or k = 2
3

• Centroid of a triangle and its coordinates: The medians of a triangle are

concurrent. Their point of concurrence is called the centroid. It divides each

median in the ratio 2 : 1. The coordinates of centroid of a triangle with vertices

A(x1, y1), B (x 2, y 2 ) and C (x 3, y 3 ) are given by

 x1 + x 2 + x 3 , y 1 + y 2 + y 3 
3 3

For example:

Find the centroid of triangle ABC where coordinates of vertices are:
A (–1, 12), B (–1, 0), C (4, 0).

Centroid of the triangle is =  − 1− 1+ 4 , 12 + 0 + 0  =  2 , 4 .
3 3  3 

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