is $ overline { a c } $ perpendicular to $ overline { x z } $ ? explain.
a. yes; the slope of $ overline { a c } $ is 1, and the slope of $ overline { x z } $ is -1.
b. yes; the slope of $ overline { a c } $ is -1, and the slope of $ overline { x z } $ is 1.
c. no; the slope of $ overline { a c } $ is 1, and the slope of $ overline { x z } $ is not -1, it is $ \frac { 2 } { 3 } $.
d. no; the slope of $ overline { a c } $ is -1, and the slope of $ overline { x z } $ is not 1, it is $ \frac { 2 } { 3 } $

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Accepted Answer

# Explanation:
## Step1: Calculate the slope of $\overline{AC}$
The formula for slope $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $A(-1,3)$ and $C(4,-2)$, $m_{AC}=\frac{-2 - 3}{4-(-1)}=\frac{-5}{5}=-1$.
## Step2: Calculate the slope of $\overline{XZ}$
For points $X(11,2)$ and $Z(6,-2)$, $m_{XZ}=\frac{-2 - 2}{6 - 11}=\frac{-4}{-5}=\frac{4}{5}
eq1$. Also, for perpendicular lines, if the slope of one line is $m_1$ and the slope of the other is $m_2$, then $m_1\times m_2=-1$. Since $m_{AC}=-1$ and $m_{XZ}
eq1$ (because $-1\times1=-1$ is the condition for perpendicularity when one slope is $-1$ and the other is $1$), we check the options.

# Answer:
D. No; the slope of $\overline{AC}$ is $-1$, and the slope of $\overline{XZ}$ is not $1$, it is $\frac{2}{3}$ (Note: There was a calculation error in the slope of $XZ$ in the step - by - step above. Redoing the slope of $XZ$ with $X(11,2)$ and $Z(6,-2)$: $m=\frac{-2 - 2}{6 - 11}=\frac{-4}{-5}=\frac{4}{5}$. But if we assume a mis - read of coordinates (maybe $X(10,2)$ and $Z(6,-2)$), $m=\frac{-2 - 2}{6 - 10}=\frac{-4}{-4} = 1$ was wrong. If we assume $X(10,2)$ and $Z(6,-2)$ was a typo and we use the general rule: Two lines are perpendicular if $m_1\times m_2=-1$. The slope of $AC$ (using $A(-1,3)$ and $C(4,-2)$) is $m_{AC}=\frac{-2 - 3}{4+1}=-1$. For $XZ$ (assuming correct coordinates from the graph: if $X(11,2)$ and $Z(6,-2)$, $m=\frac{-2 - 2}{6 - 11}=\frac{-4}{-5}=\frac{4}{5}$. But if we go by the options' logic of slope calculation (maybe wrong coordinate reading in the problem - maker's side for $XZ$ as if $X(10,2)$ and $Z(6,-2)$ was intended as $m=\frac{-2-2}{6 - 10}=1$ was wrong. But according to the formula $m = \frac{y_2-y_1}{x_2 - x_1}$, for $A(-1,3)$ and $C(4,-2)$, $m=-1$. For $XZ$ (if we assume the problem - maker made a mistake in coordinate - based slope calculation for $XZ$ as per options, but following the rule that for perpendicularity $m_1\times m_2=-1$. Since $m_{AC}=-1$, $m_{XZ}$ should be $1$ for perpendicularity. But $m_{XZ}$ (correctly calculated with $X(11,2)$ and $Z(6,-2)$) is $\frac{4}{5}$, but among the options, option D says $m_{AC}=-1$ and $m_{XZ}
eq1$)

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Question

Explanation:

Step1: Calculate the slope of \(\overline{AC}\)

Step2: Calculate the slope of \(\overline{XZ}\)

Answer: