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

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

$x=3+2 \sqrt{2} \cos \theta \quad\left(\because \tan \theta=1 \text {, then } \sin \theta=\frac{1}{\sqrt{2}}, \cos \theta=\frac{1}{\sqrt{2}}\right)$
$y=7+2 \sqrt{2} \sin \theta$
$x =3+2=5 \quad ; \quad y =9 \Rightarrow C (5,9) \quad \Rightarrow \quad A (1,5)$
$\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 $y_2=9 \Rightarrow y_2-y_1=4$