Question

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

Answer 1

$m(\frac{5+1}{2},\frac{3+1}{2})$

$\therefore m(3,2)$

equation of ${}^{\prime }B{D}^{\prime }$ is

$y=2x+c$

$\Rightarrow 2=6+c$

$\therefore c=-4$

let $(\alpha ,\beta )$

$\beta =2\alpha -4$ ....(1)

slope=$2$

$\Rightarrow \frac{\beta -3}{\alpha -1}\times \frac{\beta -1}{\alpha -5}=-1$

$\Rightarrow (2\alpha -4-3)(2\alpha -4-1)=-(\alpha -1)(\alpha -5)$

$\Rightarrow (2\alpha -7)(2\alpha -5)=-(\alpha -1)(\alpha -5)$

$\Rightarrow 4{\alpha }^2-10\alpha -14\alpha +35=-{\alpha }^2+5\alpha +2=5$

$\Rightarrow 5{\alpha }^2-30\alpha +40=0$

$\Rightarrow {\alpha }^2-6\alpha +8=0$

$\Rightarrow {\alpha }^2-4\alpha -2\alpha +8=0$

$\Rightarrow \alpha (\alpha -4)-2(\alpha -4)=0$

$\therefore \alpha =2,\alpha =4$

when $\alpha =2$, $\beta =0$

when $\alpha =4$, $\beta =4$

$\therefore B(2,0)$ & $D(4,4)$