FINDING THE EQUATION OF THE LINE

Example 1 :

Find the value of ‘a’, if the line through (–2, 3) and (8, 5) is perpendicular to y = ax +2

Solution :

Equation of the line passing through the points (-2, 3) and (8, 5).

(y - y1)/(y2 - y1)  =  (x - x1)/(x2 - x1)

(y - 3)/(5 - 3)  =  (x + 2)/(8 + 2)

(y - 3)/2  =  (x + 2)/10

10(y - 3)  =  2(x + 2)

10y - 30  =  2x + 4

2x + 10y - 30 - 4  =  0

2x + 10y - 34  =  0

Dividing the entire equation by 2, we get

x + 5y - 17  =  0

Example 2 :

The hill is in the form of a triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45˚. Find the equation of the hill joining the foot and top.

Solution :

Equation of the hill joining the foot and top :

slope (m)  =  tan 45  =  1

y - y1  =  m(x - x1)

y - 3  =  1(x - 19)

y - 3  =  x - 19

x + y - 19 + 3  =  0

x + y - 16  =  0

Example 3 :

Find the equation of a line through the given pair of points

(i)  (2, 2/3) and (-1/2, -2)

Solution :

(y - y1)/(y- y1)  =  (x - x1)/(x- x1)

(y - (2/3))/(-2 - (2/3))  =  (x - 2)/((-1/2) - 2)

((3y - 2)/3)/(-8/3)  =  (x - 2)/(-5/2)

-(3y - 2)/8  =  -2(x - 2)/5

5(3y - 2)  =  16(x - 2)

15y - 10  =  16x - 32

16x - 15y - 32 + 10  =  0 

16x - 15y - 22  =  0

(ii) (2, 3) and (-7,-1)

(y - 3)/(-1 - 3)  =  (x - 2)/(-7 - 2)

(y - 3)/(-4)  =  (x - 2)/(-9)

-9(y - 3)  =  -4(x - 2)

-9y + 27  =  -4x + 8

4x - 9y + 27 - 8  =  0

4x - 9y + 19  =  0 

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