P(4,2) and R(-2,0) are vertices of a rhombus PQRS. What is the equation of diagonal QS.

Diagonals of a rhombus bisect each other perpendicularly.

=> QS perpendicularly bisects diagonal joining  P(4,2) and R(-2,0) at O, thus O is the mid point of PR and QS.

=> Coordinates of O = $$(\frac{4 - 2}{2} , \frac{2 + 0}{2})$$

= $$(\frac{2}{2} , \frac{2}{2}) = (1,1)$$

Now, slope of PR = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2)}{(-2 - 4)}$$

= $$\frac{-2}{-6} = \frac{1}{3}$$

Let slope of line QS = $$m$$

Product of slopes of two perpendicular lines = -1

=> $$m \times \frac{1}{3} = -1$$

=> $$m = -3$$

Equation of a line passing through point $$(x_1,y_1)$$ and having slope $$m$$ is $$(y - y_1) = m(x - x_1)$$

$$\therefore$$ Equation of line QS passing thorugh O(1,1)

=> $$(y - 1) = -3(x - 1)$$

=> $$y -1 = -3x + 3$$

=> $$3x + y = 4$$

=> Ans - (B)