Question

$ABCD$ is a rectangle in the clockwise direction. The coordinates of $A$ are $(1,3)$ and of $C$ are $(5,1)$. Coordinates of vertices $B$ and $D$ satisfy $y=2x+c.$ The coordinates of the

Answer 1

Let M be the point of intersection of the diagonals. M bisects each of the diagonals.
$\therefore$ The coordinates of the middle point M of
$BD=(\frac{5+1}{2},\frac{1+3}{2})=(3,2)$
Next, as M satisfies $y=2x+c\Rightarrow 2=6+c\Rightarrow c=-4$.
Let the coordinates of D which satisfies
$y=2x-4$ be $(x,2x-4)$.
Then, from $(AD{)}^2+(DC{)}^2=(AC{)}^2$, we get
$(x-1{)}^2+(2x-4-3{)}^2+(x-5{)}^2+(2x-4-1{)}^2=(5-1{)}^2+(1-3{)}^2$
$\Rightarrow x^2-6x+8=0\Rightarrow x=2orx=4$
When $x=2,y=0$ and from the figure , as ABCD is in the clockwise direction.
We get coordinates of D as $(2,0)$.
When $x=4,y=4$ so the coordinates of B are $(4,4)$ and the coordinates of the middle point of AB are $(\frac{4+1}{2},\frac{4+3}{2})=(\frac{5}{2},\frac{7}{2})$.
The coordinates of the middle point of BC are $(\frac{4+5}{2},\frac{1+4}{2})=(\frac{9}{2},\frac{5}{2})$.
Hence, A - 2 ,B-3 , C-4 , D-1