Question

In the coordinate plane, draw ABCD with A(-5,0), B(1,-7), C(8,-1), and D(2,6). Then give the most precise name for ABCD and prove your conclusion.

Answer 1

Drawing the points A, B, C and D in the coordinate plane, we have:

In order to find the most precise name of ABCD, first let's find the slope of each side, using the formula:

`m=(y_2-y_1)/(x_2-x_1)`

So calculating the slope of each side, we have:

`\begin{gathered} A\text{ and }B\colon \\ m=(-7-0)/(1-(-5))=-(7)/(6) \\ B\text{ and }C\colon \\ m=(-1-(-7))/(8-1)=(6)/(7) \\ C\text{ and }D\colon \\ m=(6-(-1))/(2-8)=-(7)/(6) \\ D\text{ and }A\colon \\ m=\frac{0-6}{-5-2_{}}=(6)/(7) \end{gathered}`

Since the slopes of opposite sides are the same, they are parallel. Also, since the slopes of adjacent sides follow the rule m1 = -1 / m2, they are perpendicular.

So we already know that the figure is a rectangle. In order to find if the figure is a square, let's find the length of all sides using the formula for the distance between two points:

`\begin{gathered} d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2_{}} \\ d_(AB)=\sqrt[]{(-7-0)^2+(1-(-5))^2}=\sqrt[]{49+36}=\sqrt[]{85} \\ d_(BC)=\sqrt[]{(-1-(-7))^2+(8-1)^2}=\sqrt[]{36+49}=\sqrt[]{85} \\ d_(CD)=\sqrt[]{(6-(-1))^2+(2-8)^2_{}}=\sqrt[]{49+36}=\sqrt[]{85} \\ d_(DA)=\sqrt[]{(0-6)^2+(-5-2)^2}=\sqrt[]{36+49}=\sqrt[]{85} \end{gathered}`

All sides have the same length, so the figure ABCD is a square.