Question

For a square with centre at $(3,7),$ side length as $4$ units and having one of its diagonals parallel to $y=x$ the sum of the abscissa and ordinates of the vertices are

Answer 1

Slope of $AC=1$
$\Rightarrow$ Slope of $BD=-1$
$\because$ Side length $=4,$
Diagonal length $=4\sqrt{2}$
Now using parametric form,
$1)$ Coordinates at $C,A:$
$=(3\pm 2\sqrt{2}\cos \frac{\pi }{4},7\pm 2\sqrt{2}\sin \frac{\pi }{4})$
$=(3\pm 2,7\pm 2)=(5,9);(1,5)$

$2)$ Coordinates of $D,B:$
$=(3\pm 2\sqrt{2}\cos \frac{3\pi }{4},7\pm 2\sqrt{2}\sin \frac{3\pi }{4})$
$=(3\mp 2,7\pm 2)$
$=(1,9);(5,5)$

$\therefore$ Sum of ordinates and abscissa of the vertices $=40$

Alternative solution:
As diagonals are parallel to the lines $y=x$ and $y=-x.$
Sides will be parallel to the coordinate axes.


From the diagram, $A=(3-2,7,-2)=(1,5)$
$B=(3+2,7-2)=(5,5)$
$C=(3+2,7+2)=(5,9)$
$D=(3-2,7+2)=(1,9)$

$\therefore$ Sum of ordinates and abscissa of the vertices $=40$