(d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line given by (i) bisects the join of (3, – 4) and (5, 2), the mid-point of the line joining (3, – 4) and (5, 2) lies on (i). Mid- point of join of (3, – 4) and (5, 2) = \(\big(rac{3+5}{2},rac{-4+2}{2}\big)\) = (4, –1)Putting (4, –1) in (i), we get\(rac{4}{2a}+rac{-1}{a}=1\) ⇒ \(rac{4-2}{2a}\) = 1 ⇒ \(rac{2}{2a}\) = 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.