Answer :
Answer:
Option (4)
Step-by-step explanation:
Vertices of the triangle are X(0, -4), Y(2, -3) and Z(2, -6).
Slope of XY ([tex]m_{1}[/tex]) = [tex]\frac{y-y'}{x-x'}[/tex]
= [tex]\frac{-3+4}{2-0}[/tex]
([tex]m_{1}[/tex]) = [tex]\frac{1}{2}[/tex]
Similarly, slope of XZ ([tex]m_{2}[/tex]) = [tex]\frac{-6+4}{2-0}[/tex]
([tex]m_{2}[/tex]) = (-1)
[tex]m_{1}\times m_{2}=-\frac{1}{2}[/tex]
Which should be (-1) if XY and XZ are perpendicular to each other.
Now we can say that XY and XZ are not perpendicular.
Therefore, Lydia's assertion that ΔXYZ is a right triangle is not correct.
Option (4) will be the answer.
The true statement is (d) No; the slopes of segment XY and segment XZ are not opposite reciprocals.
The coordinates are given as:
X = (0,-4)
Y = (2,-3)
Z = (2,-6)
Start by calculating the slopes of XY and XZ using:
[tex]m = \frac{y_2 - y_1}{x_2-x_1}[/tex]
So, we have:
[tex]m_{xy} = \frac{-3 + 4}{2-0}[/tex]
[tex]m_{xy} = \frac{1}{2}[/tex]
Also, we have:
[tex]m_{xz} = \frac{-6 + 4}{2-0}[/tex]
[tex]m_{xz} = \frac{-2}{2}[/tex]
[tex]m_{xz} = -1[/tex]
For the shape to be a right-angle triangle, then:
[tex]m_{xy} \times m_{xz}= -1[/tex]
So, we have:
[tex]\frac 12 \times -1= -1[/tex]
[tex]-\frac 12 = -1[/tex]
The above equation is false.
Hence, the triangle is not a right-angled triangle
Read more about right-angled triangles at:
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