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 = `|(4x_1 + 3y_1 - 10)/sqrt(4^2 + 3^2)|`

∴ 1 = `|(4x_1 + 3(4 - x_1) - 10)/sqrt(25)|`  ...[From (i)]

∴ 1 = `|(4x_1 + 12 - 3x_1 - 10)/5|`

∴ 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).