To determine if the triangle [tex]\(\Delta XYZ\)[/tex] is a right triangle, we need to check the slopes of the lines [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex]. For [tex]\(\Delta XYZ\)[/tex] to be a right triangle at [tex]\(X\)[/tex], the lines [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex] should be perpendicular. This happens if the slopes of these lines are opposite reciprocals.
We are given the coordinates of the points:
- [tex]\(X(0, -4)\)[/tex]
- [tex]\(Y(2, -3)\)[/tex]
- [tex]\(Z(2, -6)\)[/tex]
First, we calculate the slope of line [tex]\(\overline{XY}\)[/tex]:
[tex]\[
\text{slope}_{XY} = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{-3 - (-4)}{2 - 0} = \frac{-3 + 4}{2 - 0} = \frac{1}{2}
\][/tex]
Next, we calculate the slope of line [tex]\(\overline{XZ}\)[/tex]:
[tex]\[
\text{slope}_{XZ} = \frac{Z_2 - Z_1}{X_2 - X_1} = \frac{-6 - (-4)}{2 - 0} = \frac{-6 + 4}{2 - 0} = \frac{-2}{2} = -1
\][/tex]
To check if [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex] are perpendicular, we need to see if the product of the slopes is [tex]\(-1\)[/tex]:
[tex]\[
\text{slope}_{XY} \times \text{slope}_{XZ} = \frac{1}{2} \times (-1) = -\frac{1}{2}
\][/tex]
The product of the slopes [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-1\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex], which is not [tex]\(-1\)[/tex]. Therefore, the lines [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex] are not perpendicular.
Thus, Lydia's assertion is not correct. Therefore, the correct answer is:
No; the slopes of [tex]\(\overline{XY}\)[/tex] and [tex]\(\overline{XZ}\)[/tex] are not opposite reciprocals.
