Question

Find points on the line x + y − 4 = 0 which are at one unit distance from the line 4x + 3y – 10 = 0.

Answer 1

Let P(x₁, y₁) be a point on the line x + y − 4 = 0.

∴ x₁ + y₁ − 4 = 0

∴ y₁ = 4 − x₁   ...(i)

Also, distance of P from the line 4x + 3y – 10 = 0 is 1

∴ 1 = ${\displaystyle \left|\frac{4x_1+3y_1-10}{\sqrt{4^2+3^2}}\right|}$

∴ 1 = ${\displaystyle \left|\frac{4x_1+3(4-x_1)-10}{\sqrt{25}}\right|}$  ...[From (i)]

∴ 1 = ${\displaystyle \left|\frac{4x_1+12-3x_1-10}{5}\right|}$

∴ 5 = |x₁ + 2|

∴ x₁ + 2 = ± 5

∴ x₁ + 2 = 5 or x₁ + 2 = − 5

∴ x₁ = 3 or x₁ = − 7

From (i), when x₁ = 3, y₁ = 1

and when x₁ = −7, y₁ = 11

∴ The required points are (3, 1) and (−7, 11).