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Alenkasestr [34]
1 year ago
6

Consider AABC with vertices A(-2, -3) and B(-6, -6) and C(-8, 5).a) Draw AABC on graph paper. Is this a right triangle? JUSTIFY

your answer.b) Reflect AABC across AC to create AA'B'C'. Find the location of B'. What kind oftriangle is ABB'C:? JUSTIFY your answer.
Mathematics
1 answer:
Fiesta28 [93]1 year ago
6 0

Given data:

The graph of the given points is shown below.

The slope of the line passing through A(-2, -3) and B(-6, -6).

\begin{gathered} m=\frac{-6-(-3)}{-6-(-2)} \\ =\frac{-3}{-4} \\ =\frac{3}{4} \end{gathered}

The slope of the line passing through B(-6, -6) and C(-8, 5).

\begin{gathered} m^{\prime}=\frac{5-(-6)}{-8-(-6)} \\ =-\frac{11}{2} \end{gathered}

The slope of the line passing through A(-2, -3) and C(-8, 5).

\begin{gathered} m^{\doubleprime}=\frac{5-(-3)}{-8-(-2)} \\ =\frac{8}{-6} \\ =-\frac{4}{3} \end{gathered}

The product of the slope of line passing through AB and line passing through AC is,

\begin{gathered} m\times m^{\doubleprime}=\frac{3}{4}\times(-\frac{4}{3}_{}) \\ =-1 \end{gathered}

Thus, the line passing through AB and line passing through AC are perpendicular.

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