Question

Each side of a square is of length 4 . The centre of the square is $(3,7)$ and one of its diagonals is parallel to $y=x$. If the smallest ordinate of all the vertices is $y_1$ and the largest ordinate is $y_2$ then find the value of $\left(y_2-y_1\right)$.

Answer 1

Answer: 4

Let line $AC$ is parallel to $y=x$
$A{C}^2=16+16=32$
$AC=4\sqrt{2}$
$\frac{x-3}{\cos \theta }=\frac{y-7}{\sin \theta }=\pm 2\sqrt{2}$

$x=3+2\sqrt{2}\cos \theta \left(\because \tan \theta =1\text{, then }\sin \theta =\frac{1}{\sqrt{2}},\cos \theta =\frac{1}{\sqrt{2}}\right)$
$y=7+2\sqrt{2}\sin \theta$
$x=3+2=5;y=9\Rightarrow C(5,9)\Rightarrow A(1,5)$
$\text{ again }\frac{x-3}{\cos \alpha }=\frac{y-7}{\sin \alpha }=2\sqrt{2}\text{ where }\cos \alpha =-\frac{1}{\sqrt{2}};\sin \alpha =\frac{1}{\sqrt{2}}$
$x=5;y=-2+7=5$
Then $B(5,5)$ and $D(1,9)$
Coordinates are $A(1,5);B(5,5);C(5,9);D(1,9)$
Hence $y_1=5$ and $y_2=9\Rightarrow y_2-y_1=4$