Answer
The points $(-2,9)$, $(4,6)$ and $(1,0)$, $(-5,3)$ are vertices of a square.
Work Step by Step
We are given the square with vertices $A(-2,9)$, $B(4,6)$ and $C(1,0)$, $D(-5,3)$.
Consider AB, BC , CD and AD are four sides of a square.
Use distance formula $d=\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
and slope formula $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
For side AB:
$AB=\sqrt {(4-(-2))^{2}+(6-9)^{2}}=\sqrt {36+9}=\sqrt {45}$
$m=\frac{6-9}{4-(-2)}=-\frac{1}{2}$
For side BC:
$BC=\sqrt {(1-4)^{2}+(0-6)^{2}}=\sqrt {9+36}=\sqrt {45}$
$m=\frac{0-6}{1-4}=2$
For side CD:
$CD=\sqrt {(-5-1)^{2}+(3-0)^{2}}=\sqrt {36+9}=\sqrt {45}$
$m=\frac{3-0}{-5-1}=-\frac{1}{2}$
For side AD:
$AD=\sqrt {(-5-(-2))^{2}+(3-9)^{2}}=\sqrt {9+36}=\sqrt {45}$
$m=\frac{3-9}{-5-(-2)}=2$
All sides are equal and their consecutive sides are perpendicular having negative reciprocal of their slope, therefore, the given points are vertices of a square.
Hence, the points $(-2,9)$, $(4,6)$ and $(1,0)$, $(-5,3)$ are vertices of a square.
