Question

A rectangle $ABCD$ has its side $AB$ parallel to line $y=x$ and vertices $A,B$ and $D$ lie on $y=1,x=2$ and $x=-2$, respectively. Locus of vertex ${}^{\prime }{C}^{\prime }$ is

• A. $x=5$
• B. $x-y=5$
• C. $y=5$
• D. $x+y=5$

Answer 1

Answer: (C)

$y=5$

Let the equation of side $BC$ be $y=x+a$
$\Rightarrow A=(1-a,1),B=(2,2+a)$
Equation of side $AD$ is
$y-1=-[x-(1-a\left)\right]$
$\Rightarrow D\equiv (-2,4-a)$
Let $C\equiv (h,k)\Rightarrow h+1-a=2-2$
$\Rightarrow h=a-1$ and $k+1=2+a+4-a\Rightarrow k=5$
Thus, the locus of $C$ is $y=5$.