Question

1. a triangle has vertices d(-2, 7), e(-4, 2), and f(6, -2).

a) show algebraically that this triangle is a right triangle.
b) find the midpoint of the hypotenuse.
c) show that this midpoint is equidistant from each of the vertices.

Explanation:

Step1: Calculate slopes of sides

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $DE$: $m_{DE}=\frac{2 - 7}{-4+2}=\frac{-5}{-2}=\frac{5}{2}$.
Slope of $EF$: $m_{EF}=\frac{-2 - 2}{6 + 4}=\frac{-4}{10}=-\frac{2}{5}$.
Slope of $DF$: $m_{DF}=\frac{-2 - 7}{6+2}=\frac{-9}{8}$.
Since $m_{DE}\times m_{EF}=\frac{5}{2}\times(-\frac{2}{5})=- 1$, $DE\perp EF$, so $\triangle DEF$ is a right - triangle.

Step2: Identify hypotenuse and find mid - point

The hypotenuse is the side opposite the right angle. Here, the hypotenuse is $DF$.
The mid - point formula is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.
For $D(-2,7)$ and $F(6,-2)$, the mid - point $M$ of $DF$ is $(\frac{-2 + 6}{2},\frac{7-2}{2})=(2,\frac{5}{2})$.

Step3: Calculate distances from mid - point to vertices

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Distance from $M(2,\frac{5}{2})$ to $D(-2,7)$:
\[

$$\begin{align*} d_{MD}&=\sqrt{(2 + 2)^2+(\frac{5}{2}-7)^2}\\ &=\sqrt{16+(\frac{5 - 14}{2})^2}\\ &=\sqrt{16+\frac{81}{4}}\\ &=\sqrt{\frac{64 + 81}{4}}\\ &=\sqrt{\frac{145}{4}} \end{align*}$$

\]
Distance from $M(2,\frac{5}{2})$ to $E(-4,2)$:
\[

$$\begin{align*} d_{ME}&=\sqrt{(2 + 4)^2+(\frac{5}{2}-2)^2}\\ &=\sqrt{36+(\frac{5 - 4}{2})^2}\\ &=\sqrt{36+\frac{1}{4}}\\ &=\sqrt{\frac{144+1}{4}}\\ &=\sqrt{\frac{145}{4}} \end{align*}$$

\]
Distance from $M(2,\frac{5}{2})$ to $F(6,-2)$:
\[

$$\begin{align*} d_{MF}&=\sqrt{(6 - 2)^2+(-2-\frac{5}{2})^2}\\ &=\sqrt{16+(-\frac{4 + 5}{2})^2}\\ &=\sqrt{16+\frac{81}{4}}\\ &=\sqrt{\frac{64 + 81}{4}}\\ &=\sqrt{\frac{145}{4}} \end{align*}$$

\]
Since $d_{MD}=d_{ME}=d_{MF}=\sqrt{\frac{145}{4}}$, the mid - point of the hypotenuse is equidistant from each of the vertices.

Answer:

a) By showing the product of the slopes of $DE$ and $EF$ is $-1$, the triangle is a right - triangle.
b) The mid - point of the hypotenuse $DF$ is $(2,\frac{5}{2})$.
c) By calculating the distances from the mid - point $(2,\frac{5}{2})$ to each vertex and showing they are equal ($\sqrt{\frac{145}{4}}$), the mid - point is equidistant from each of the vertices.