Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

Let (x₁,y₁) be any point lying in the equation x+ y = 4

∴ x₁ + y₁ = 4 …(i)

Distance of the point (x₁,y₁) from the equation 4x + 3y = 10

[given]

⇒ 4x₁ + 3y₁ – 10 = ±5

either 4x₁ + 3y₁ – 10 = 5 or 4x₁ + 3y₁ – 10 = -5

4x₁ + 3y₁ = 5 + 10 or 4x₁ + 3y₁ = -5 + 10

4x₁ + 3y₁ = 15 …(ii) or 4x₁ + 3y₁ = 5 …(iii)

From eq. (i), we have y₁ = 4 – x₁ …(iv)

Putting the value of y₁ in eq. (ii), we get

4x₁ + 3(4 – x₁) = 15

⇒ 4x₁ + 12 – 3x₁ = 15

⇒ x₁ = 15 – 12

⇒ x₁ = 3

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – 3

⇒ y₁ = 1

Putting the value of y₁ = 4 – x₁ in eq. (iii), we get

4x₁ + 3(4 – x₁) = 5

⇒ 4x₁ + 12 – 3x₁ = 5

⇒ x₁ = 5 – 12

⇒ x₁ = - 7

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – (-7)

⇒ y₁ = 4 + 7

⇒ y₁ = 11

Hence, the required points on the given line are (3,1) and (-7,11)