Question

Find the distance of a straight line which passes through the point (a, 0) and whose perpendicular distance from (2a, 2a) is a.

Answer 1

Let the straight line be Ax + By + C = 0      (1)

Since (1) passes through (a, 0), so

A. a + B. 0 + C = 0

⇒ Aa + C = 0

Also, the ⊥ distance of (1) from (2a, 2a) is a.

⇒ A² + B² = A² + 4AB + 4B²

⇒ – 3B² = 4 AB

⇒ 4AB + 3B² = 0

⇒ B(4A + 3B) = 0

⇒ B = 0 or 4A + 3B = 0

If 4A + 3B = 0

⇒ B = − \(\frac{4A}{3}\)

From eq. (i)

Ax = − \(\frac{4}{3}\) Ay − Aa = 0

∴ Equation of a line is 3x – 4y – 3a = 0

If B = 0 then equation of the line is x – a = 0