Question

Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2x – 3y + 4 = 0 and making intercept 3 on the X-axis.

Answer 1

Given equations of lines are
x + y – 2 = 0                 ...(i)
and 2x – 3y + 4 = 0     ...(ii)
Multiplying equation (i) by 3, we get
3x + 3y – 6 = 0           ...(iii)
Adding equation (ii) and (iii), we get
5x – 2 = 0

∴ x = `2/5`

Substituting x = `2/5` in equation (i), we get

`2/5 + y - 2` = 0

∴ y = `2 - 2/5 = 8/5`

∴ The required line passes through point `(2/5, 8/5)`.

Also, the line makes intercept of 3 on X-axis
∴ it also passes through point (3, 0).
∴ required equation of line passing through points `(2/5, 8/5)` and (3, 0) is

`(y - 8/5)/(0 - 8/5) = (x - 2/5)/(3 - 2/5)`

∴ `((5y - 8)/5)/(-8/5) = ((5x - 2)/5)/(13/5)`

∴ `(5y - 8)/(-8) = (5x - 2)/13`

∴ 13 (5y – 8) = – 8 (5x – 2)
∴ 65y – 104 = – 40x + 16
∴ 40x + 65y – 120 = 0
∴ 8x + 13y – 24 = 0 which is the equation of the required line.