Lydia graphed [tex]$\triangle XYZ$[/tex] at the coordinates [tex]$X(0,-4)$[/tex], [tex]$Y(2,-3)$[/tex], and [tex]$Z(2,-6)$[/tex]. She thinks [tex]$\triangle XYZ$[/tex] is a right triangle. Is Lydia's assertion correct?

A. Yes; the slopes of [tex]$\overline{XY}$[/tex] and [tex]$\overline{XZ}$[/tex] are the same.
B. Yes; the slopes of [tex]$\overline{XY}$[/tex] and [tex]$\overline{XZ}$[/tex] are opposite reciprocals.
C. No; the slopes of [tex]$\overline{XY}$[/tex] and [tex]$\overline{XZ}$[/tex] are not the same.
D. No; the slopes of [tex]$\overline{XY}$[/tex] and [tex]$\overline{XZ}$[/tex] are not opposite reciprocals.



Answer :

To determine if [tex]\(\triangle XYZ\)[/tex] is a right triangle, we need to analyze the slopes of the sides to see if any pair of slopes are opposite reciprocals. Opposite reciprocal slopes indicate perpendicular lines, which is a requirement for one of the angles to be 90 degrees in a right triangle.

Let's find the slopes of each pair of points.

1. Slope of [tex]\(\overline{XY}\)[/tex]:
- Points [tex]\(X(0, -4)\)[/tex] and [tex]\(Y(2, -3)\)[/tex]
- Slope formula: [tex]\(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex]
- [tex]\(\text{slope}_{XY} = \frac{-3 - (-4)}{2 - 0} = \frac{-3 + 4}{2} = \frac{1}{2}\)[/tex]

2. Slope of [tex]\(\overline{XZ}\)[/tex]:
- Points [tex]\(X(0, -4)\)[/tex] and [tex]\(Z(2, -6)\)[/tex]
- Slope formula: [tex]\(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex]
- [tex]\(\text{slope}_{XZ} = \frac{-6 - (-4)}{2 - 0} = \frac{-6 + 4}{2} = \frac{-2}{2} = -1\)[/tex]

3. Slope of [tex]\(\overline{YZ}\)[/tex]:
- Points [tex]\(Y(2, -3)\)[/tex] and [tex]\(Z(2, -6)\)[/tex]
- Slope formula: [tex]\(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex]
- Since the [tex]\(x\)[/tex] coordinates are the same (vertical line), [tex]\(\text{slope}_{YZ} = \infty\)[/tex] (indicating a vertical line)

Now, compare the slopes of the segments to check for perpendicularity (opposite reciprocals):

- Slope of [tex]\(\overline{XY} = \frac{1}{2}\)[/tex]
- Slope of [tex]\(\overline{XZ} = -1\)[/tex]
- Slope of [tex]\(\overline{YZ}\)[/tex] is vertical ([tex]\(\infty\)[/tex])

Next, we check if any pair of these slopes are opposite reciprocals. For [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-1\)[/tex], we can determine if [tex]\(\frac{1}{2} \times -1 = -\frac{1}{2}\)[/tex]. This is not equal to [tex]\(-1\)[/tex] (not opposite reciprocals).

For the vertical line [tex]\(\overline{YZ}\)[/tex], we consider it’s perpendicular to any horizontal line which means it has slopes [tex]\(\infty\)[/tex] versus [tex]\(0\)[/tex]. But there is no horizontal line here.

Thus, none of the slopes are opposite reciprocals indicating perpendicularity, hence not right angle formed.

Therefore, the answer is:

No; the slopes of [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex] are not opposite reciprocals.