Question

Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18, 4x + 5y = 9 and bisecting the line segment joining the points (5, −4) and (−7, 6)

Answer 1

Given lines are

                                             8x + 3y = 18    ...(1)
                                             4x + 5y = 9      ...(2)
(1) × 5 ⇒                          40x + 15y = 90    ...(3)
(2) × 3 ⇒                          12x + 15y = 27    ...(4)
                                          (−)   (−)       (−) 
Subtracting (3) and (4) ⇒ 28x  +    0 = 63

x = `63/28 = 9/4`

Substitute the value of x = `9/4` in (2)

`4(9/4) + 5y` = 9

9 + 5y = 9

⇒ 5y = 9 – 9

5y = 0

⇒ y = 0

The point of intersection is `(9/4, 0)`

Mid point of the points (5, – 4) and (– 7, 6)

Mid point of a line = `((x_1 +x_2)/2, (y_1 + y_2)/2)`

= `((5 - 7)/2, (-4 + 6)/2)`

= `(-2/2, 2/2)`

= (– 1, 1)

Equation of the line joining the points `(9/4, 0)` and (– 1, 1)

`(y - y_1)/(y_2 - y_1) = (x - x_1)/(x_2 - x_1)`

`(y-0)/(1 - 0) = (x - 9/4)/(-1 -9/4)`

⇒ y = `((4x - 9)/4)/((-4 - 9)/4)`

= `(4x - 9)/4 xx 4/(-13)`

y = `(4x - 9)/(-13)`

– 13y = 4x – 9

– 4x – 13y + 9 = 0

⇒ 4x + 13y – 9 = 0

The equation of the line is 4x + 13y – 9 = 0