Question

Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Answer 1

∵ x-intercept = 3

∴ The line passes through A(3, 0).

Line PQ: x – 7y + 5 = 0

or 7y = x + 5

or y = `1/7 "x" + 5/7`

Therefore slope of PQ = `1/7`

∵ PQ ⊥ AB

∴ Slope of line AB passing through A = –7

∴ Equation of line AB from point (3, 0),

y – 0 = –7(x – 3)

= –7x + 21

or 7x + y – 21 = 0

Second method: Any line perpendicular to ax + by + c = 0 bx – ay + k = 0

∴ Perpendicular to x – 7y + 5 = 0 7x + y + k = 0

This line passes through (3, 0).

∴ 7 x 3 + 0 + k = 0,

i.e. k = –21

∴ The equation of the required line is 7x + y – 21 = 0.