Question

A rectangle $ABCD$ has its side $AB$ parallel to line $y=x$ and the vertices $A,B,D$ lie on lines $y=1,$ $x=2,x=-2$, respectively. Then the locus of the vertex $C$ is

• A. $x-y=5$
• B. $x+y=3$
• C. $x=3$
• D. $y=5$

Answer 1

The correct option is D $y=5$

Let the coordinates of $C$ be $(h,k)$
The $x$-coordinate of $D$ is $-2$.
Using slope of $CD=1$, we get $y$-coordinate of $D$ as $k-h-2$
Similarly, $x$-coordinate of $B$ is $2$.
Using slope of $BC=-1$, we get Y-coordinate of $B$ as $h+k-2$
Similarly, $y$-coordinate of $A$ is $1$.
Using slope of $AB=1$, we get $x$-coordinate of $A$ as $5-h-k$
Since the diagonals of a rectangle bisect each other, midpoints of $AC$ and $BD$ coincide.
Midpoint of segment $AC:(\frac{5-k}{2},\frac{1+k}{2})$
Midpoint of segment $BD:(0,\frac{2k-4}{2})$
Equating coordinates , we get $k=5$
Hence, locus of point $C$ is $y=5$