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

The correct option is C $y=5$

Let the equation of line $AB$ is $y=x+c$
vertices of $A,B$ are $(1-c,1)(2,2+c)$
As, line $AD$ is perependicular to line $AB$ so slope of $AD=-1$
Equation of line $AD$ is $(y-1)=-1(x-(1-c\left)\right)$
As point $D$ lie on $x=-2$
So, $y-1=-1(-2-(1-c\left)\right)\Rightarrow y=4-c$
$\therefore$ point $D$ is $(-2,4-c)$
Let point $C$ is $(h,k)$

y-coordinate of mid point $M$ is
$\frac{k+1}{2}=\frac{2+c+4-c}{2}$
$\Rightarrow k=5$
$\Rightarrow y=5$