Question

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

• A. $c=4;B(2,0);D(4,4)$
• B. $c=-4;B(2,0);D(4,4)$
• C. $c=-4;B(-2,0);D(-4,2)$
• D. $c=4;B(2,0);D(-4,-4)$

Answer 1

The correct option is B $c=-4;B(2,0);D(4,4)$

Since, diagonals of rectangle bisect each other, so mid point of $(1,3)$ and $(5,1)$ must satisfy $y=2x+c$ i.e $(3,2)$ lies on it

$\Rightarrow 2=6+c\Rightarrow c=-4$

Therefore other two vertices lies on $y=2x-4$

Let the coordinate of $B$ be $(x,2x-4)$

Therefore slope of $AB$ . slope of $BC=-1$

$\Rightarrow \left(\frac{2x-4-3}{x-1}\right).\left(\frac{2x-4-1}{x-5}\right)=-1$

$\Rightarrow (x^2-6x+8)=0\Rightarrow x=4,2\Rightarrow y=4,0$

Hence, required points are $(4,4),(2,0)$