Question

which statement proves that △xyz is an isosceles right triangle? the slope of $overline{xz}$ is $\frac{3}{4}$, the slope of $overline{xy}$ is $-\frac{4}{3}$, and the slope of $overline{zy}=7$. $overline{xz}perpoverline{xy}$ the slope of $overline{xz}$ is $\frac{3}{4}$, the slope of $overline{xy}$ is $-\frac{4}{3}$, and $xz = xy = 5$. $xz = xy = 5$

Explanation:

Step1: Recall properties of isosceles right - triangle

An isosceles right - triangle has two equal sides and one right angle.

Step2: Analyze slope condition for right - angle

If the product of the slopes of two lines is - 1, the lines are perpendicular. For two lines with slopes $m_1$ and $m_2$, $m_1\times m_2=-1$ means perpendicularity. Here, if the slope of $\overline{XZ}$ is $\frac{3}{4}$ and the slope of $\overline{XY}$ is $-\frac{4}{3}$, then $\frac{3}{4}\times(-\frac{4}{3})=- 1$, so $\overline{XZ}\perp\overline{XY}$, which gives a right - angle at $X$.

Step3: Analyze side - length condition for isosceles

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For points $X(1,3)$ and $Z(5,6)$:
\[

$$\begin{align*} XZ&=\sqrt{(5 - 1)^2+(6 - 3)^2}\\ &=\sqrt{4^2+3^2}\\ &=\sqrt{16 + 9}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]
For points $X(1,3)$ and $Y(4,-1)$:
\[

$$\begin{align*} XY&=\sqrt{(4 - 1)^2+(-1 - 3)^2}\\ &=\sqrt{3^2+(-4)^2}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]
Since $XZ = XY = 5$ and $\overline{XZ}\perp\overline{XY}$, $\triangle XYZ$ is an isosceles right - triangle.

Answer:

The slope of $\overline{XZ}$ is $\frac{3}{4}$, the slope of $\overline{XY}$ is $-\frac{4}{3}$, and $XZ = XY = 5$.