Question

three vertices of a square are (-5, -2), (1, 5), and (8, -1). graph the square.

Explanation:

Step1: Label given vertices

Let $A=(-5,-2)$, $B=(1,5)$, $C=(8,-1)$.

Step2: Calculate side lengths

Use distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$:
$AB=\sqrt{(1-(-5))^2+(5-(-2))^2}=\sqrt{6^2+7^2}=\sqrt{36+49}=\sqrt{85}$
$BC=\sqrt{(8-1)^2+(-1-5)^2}=\sqrt{7^2+(-6)^2}=\sqrt{49+36}=\sqrt{85}$
$AC=\sqrt{(8-(-5))^2+(-1-(-2))^2}=\sqrt{13^2+1^2}=\sqrt{170}$
Since $AB^2+BC^2=85+85=170=AC^2$, $AB\perp BC$, so $AB$ and $BC$ are adjacent sides.

Step3: Find 4th vertex $D(x,y)$

In a square, $\vec{AB}=\vec{DC}$. $\vec{AB}=(1-(-5),5-(-2))=(6,7)$.
$\vec{DC}=(8-x,-1-y)$, so:
$8-x=6 \implies x=2$
$-1-y=7 \implies y=-8$
Thus $D=(2,-8)$.

Step4: Plot and connect points

Plot $A(-5,-2)$, $B(1,5)$, $C(8,-1)$, $D(2,-8)$, then connect $A-B-C-D-A$.

Answer:

The fourth vertex of the square is $(2, -8)$. When plotting, connect $(-5, -2)$ to $(1, 5)$, $(1, 5)$ to $(8, -1)$, $(8, -1)$ to $(2, -8)$, and $(2, -8)$ back to $(-5, -2)$ to form the square.