Question

The points $(1,3)$ and $(5,1)$ are two opposite vertices of a rectangle. The other vertices lie on the line $y=2x+c$. Find $c$ and the remaining vertices ?

Answer 1

The midpoint of the given lines is: $(3,2)$.
As the diagonals of a rectangle bisect each other, $(3,2)$ must lie on the given line.
$\therefore 3=2\left(2\right)+c⟹c=-1$
Let $P$ be one of the vertex that lies on the line.
As $P$ satisfies the equation of the line, $P\equiv (h,2h-1)$
Using the fact that the sides of a rectangle are perpendicular:
$\frac{3-(2h-1)}{1-\left(h\right)}\cdot \frac{1-(2h-1)}{5-\left(h\right)}=-1$
$\frac{4-2h}{1-h}\cdot \frac{2-2h}{5-h}=-1$
$⟹(4-2h)(2-2h)+(1-h)(5-h)=0$
$⟹(1-h)\left(2\right(4-2h)+(5-h\left)\right)=0$
$⟹(1-h)(13-5h)=0$
$⟹h=1,\frac{5}{13}$
Other two vertices: $(1,1),(\frac{5}{13},\frac{-3}{13})$