(b) 3x – 2y + 5
Let the equation of the L be y = mx + c
Since O′ (–2, 6) is the image of the point O (4, 2) in line
L = 0, the mid-point of OO′, i.e., \(\bigg(\frac{-2+4}{2},\frac{6+2}{2}\bigg)\), i.e.,
(1, 4) will lie on the given line.
Also, L ⊥ OO′, so
Slope of L x Slope of OO′ = –1
⇒ m x \(\bigg(\frac{6-2}{-2-4}\bigg)\)= -1 ⇒ m = \(\frac{-1}{\frac{-2}{3}}\) = \(\frac{3}{2}\)
∴ Equation of L is y = \(\frac{3}{2}\,x+c\)
∵ It passes through (1, 4)
4 = \(\frac{3}{2}\) x 1 + c ⇒ c = 4 - \(\frac{3}{2}\) = \(\frac{5}{2}.\)
∴ Required equation is y = \(\frac{3}{2}x\) + \(\frac{5}{2}\)
⇒ 3x – 2y + 5 = 0
∴ L = 3x – 2y + 5.