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. $y=5$
• C. $x-y=5$
• D. $x+y=5$

Answer 1

The correct option is B $y=5$

Let the equation of side $AB$ be $y=x+a$.

Then, $A\equiv (1-a,1)$, $B\equiv (2,2+a)$. The equation of side $AD$ is $y-1=-\{x-(1-a\left)\right\}$.
Hence, $D\equiv (-2,4-a)$.
Let $C\equiv (h,k)$. Then, $h+1-a=2-2$ or $h=a-1$
and $k+1-a=2+a+4-a$ or $k=5$
Thus, the locus of $C$ is $y=5$.