Question

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

Answer 1

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

`(4x_1 + 3y_1 - 10)/sqrt((4)^2 + (3)^2)` = 1

`|(4x_1 + 3y_1 - 10)/5|` = 1

4x₁ + 3y₁ – 10 = ± 5

Taking (+) sign 4x₁ + 3y₁ – 10 = 5

⇒ 4x₁ + 3y₁ = 15  ......(ii)

From equation (i) we get y₁ = 4 – x₁

Putting the value of y₁ in equation (ii) we get

4x₁ + 3(4 – x₁) = 15

⇒ 4x₁ + 12 – 3x₁ = 15

⇒ x₁ + 12 = 15

⇒ x₁ = 3 and y₁ = 4 – 3 = 1

So, the required point is (3, 1)

Now taking (–) sign, we have

4x₁ + 3y₁ – 10 = – 5

⇒ 4x₁ + 3y₁ = 5   .....(iii)

From equation (i) we get y₁ = 4 – x₁

⇒ 4x₁ + 3(4 – x₁) = 5

⇒ 4x₁ + 12 – 3x₁ = 5

⇒ x₁ = 5 – 12 = – 7

and y₁ = 4 – (– 7) = 11

So, the required point is (– 7, 11)

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