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# Coordinate Geometry Summary

Let us look at one example to get the Coordinate Geometry summary that consolidates all the concepts and learnings from this topic.

Example : The diagram shows points on the number plane where side BC is parallel to x-axis. Find the coordinates of A and C

Solution: Given that OA = 3 units on x-axis, the coordinates of A is (3, 0).

C is a point on y-axis, therefore x = 0 here, and the y-coordinate of B is 4 which would be the same for C. Hence coordinates of C is (0, 4).

Therefore the coordinates of A is (3, 0) and C is (0, 4).

Find the equation of BC.

Solution: BC is parallel to x-axis, hence gradient  =  m  =  0, and it passes through (0, 4).

So equation of line BC is y = 4

Find the length of BC

Solution: Given B (8, 4) and C (0, 4)

BC  = = = = =  8 units

What is the area of ABC?

Solution: Base BC = 8 units

Height of triangle = OC  =  4 units

Area of ABC  = x BC x OC

= =  16 unit2

Find the equation of the line AB

Solution: A = (3, 0)  and B = (8, 4)

m  = = = m  = and A (3, 0), we get

(y – y1)  =  m(x – x1)

(y – 0)  = (x – 3)

5(y – 0)  =  4(x – 3)

5y  =  4x – 12,  or

4x – 5y – 12 = 0

Find the equation of the line perpendicular to AB and passing through point C (0, 4).

Solution: Equation of line AB is 4x – 5y – 12 = 0, and it has a gradient = Gradient of line perpendicular to BC is

m2  = = = So m2  = and passes through (0, 4).

Equation of the required line with m2  = and passing through (0, 4) is –

(y – y1)  =  m(x – x1)

(y – 4)  = (x – 0)

4(y – 4)  =  -5x

4y – 16  =  -5x  or

5x + 4y – 16 = 0

Find the coordinates of point of intersection of lines AB and AD, that is perpendicular to BC. i.e. Find the coordinates of D.

Solution:

BA: 4x – 5y -12 = 0  or 4x – 5y = 12     –  Eqn 1

AD: 5x + 4y -16 = 0  or 5x + 4y = 16     –  Eqn 2

Solving these simultaneous equations – 5 x Eqn 1  –  4 x Eqn 2

 20x – 25y = 60 – 20x – +16y = -64 41y = -4

y  = Substituting y  = in Eqn 1, we get

4x  –  5 x =  12

4x  – =  12

4x  =  12 + = = x  = = Therefore the coordinates of D are Show that the ares of ABC is 16 units by finding the perpendicular distance of AB from C, and using the length of AB.

Solution: Equation of line AB: 4x – 5y -12 = 0  and C: (0, 4)

Perpendicular distance formula is – d  = x1 =0, y1 = 4,  a = 4, b = -5, c = -12

d  = = = d  = units

Length of AB  = A: (3, 0)  B: (8, 4)

AB  = = = = units

Area of ABC  = x AB x d

= =  16 units2