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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.Summary example

 

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  =  sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

=  sqrt{(8~-~0)^2~+~(4~-~4)^2}

=  sqrt{8^2~+~0^2}

=  sqrt{8^2}

=  8 units

 

What is the area of DeltaABC?

Solution: Base BC = 8 units

Height of triangle = OC  =  4 units

Area of DeltaABC  =  1/2 x BC x OC

=  {8~*~4}/2

=  16 unit2

 

Find the equation of the line AB

Solution: A = (3, 0)  and B = (8, 4)

m  =  {y_2~-~y_1}/{x_2~-~x_1}

=  {4~-~0}/{8~-~3}

=  4/5

Using Point Gradient formula

m  =  4/5  and A (3, 0), we get

(y – y1)  =  m(x – x1)

(y – 0)  =  4/5(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 = 4/5

Gradient of line perpendicular to BC is

m2  =  -1/m_1

=  -1/{4/5}

=  -5/4

So m2  =  -5/4  and passes through (0, 4).

Equation of the required line with m2  =  -5/4  and passing through (0, 4) is –

(y – y1)  =  m(x – x1)

(y – 4)  =  -5/4(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  =  4/41

Substituting y  =  4/41 in Eqn 1, we get

4x  –  5 x 4/41  =  12

4x  –  20/41  =  12

4x  =  12 + 20/41

=  {~12~*~20~+~20~}/{41}

=  {~512~}/{41}

x  =  {512}/{~41~*~4~}

=   {~128~}/{41}

Therefore the coordinates of D are ({128/41},~{4/41})

 

Show that the ares of DeltaABC 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  =  {~delim{|}{ax_1~+~by_1~+~c}{|}~}/{sqrt{a^2~+~b^2}}

x1 =0, y1 = 4,  a = 4, b = -5, c = -12

d  =  {~delim{|}{4~*~0~-~5~*~4~-~12}{|}~}/{sqrt{4^2~+~(-5)^2}}

=  {~delim{|}{-20~-~12}{|}~}/{sqrt{16~+~25}}

=  {~delim{|}{-32}{|}~}/{sqrt{41}}

d  =  32/{~sqrt{41}~}  units

Length of AB  =  sqrt{(x_2~-~x_1)^2~+~(y_2~-~y_1)^2}

A: (3, 0)  B: (8, 4)

AB  =  sqrt{(8~-~3)^2~+~(4~-~0)^2}

=  sqrt{5^2~+~4^2}

=  sqrt{25~+~16}

=  sqrt{41}  units

Area of DeltaABC  =  1/2 x AB x d

=  {1/2}~*~sqrt{41}~*~32/sqrt{41}

=  16 units2