Question

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.

Answer 1

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.