P(3,1) and R(-7,5) are vertices of a rhombus PQRS. What is the equation of diagonal QS?

Diagonals of a rhombus bisect each other perpendicularly. Thus, O is the mid point of QS and RP.

Coordinates of midpoint of line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ is $$(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$$

=> $$O = (\frac{3 - 7}{2} , \frac{1 + 5}{2}) = (-2 , 3)$$

Slope of PR = $$\frac{1 - 5}{3 + 7} = \frac{-2}{5}$$

Product of slopes of two perpendicular lines is -1

=> Slope of QS = $$m \times \frac{-2}{5} = -1$$

=> $$m = \frac{5}{2}$$

$$\therefore$$ Equation of line QS with slope $$m = \frac{5}{2}$$ and passing through point O $$(x_1,y_1) = (-2,3)$$

= $$(y - y_1) = m (x - x_1)$$

= $$(y - 3) = \frac{5}{2} (x + 2)$$

= $$2y - 6 = 5x + 10$$

= $$5x - 2y = -16$$