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