## Calculus: Early Transcendentals 8th Edition

The points $(-2,9)$, $(4,6)$ and $(1,0)$, $(-5,3)$ are vertices of a square.
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. $d=\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ 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 slopes, so 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.