7)
In order to classify the quadrilateral BCDE, let's find the slope of all four sides.
The slope 'm' of two points (x1, y1) and (x2, y2) is calculated as:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
So using the four pairs of points, we have:
[tex]\begin{gathered} B(-6,5),C(6,2)\colon \\ m=\frac{2-5}{6-(-6)}=-\frac{3}{12}=-\frac{1}{4} \\ \\ C(6,2),D(2,-9)\colon \\ m=\frac{-9-2}{2-6}=-\frac{11}{-4}=\frac{11}{4} \\ \\ D(2,-9),E(-10,-6)\colon_{} \\ m=\frac{-6-(-9)}{-10-2}=\frac{3}{-12}=-\frac{1}{4} \\ \\ E(-10,-6),B(-6,5)\colon \\ m=\frac{5-(-6)}{-6-(-10)}=\frac{11}{4} \end{gathered}[/tex]
We can see that we have two pairs of parallel slopes, so we have a parallelogram.
In order to find if the angles are 90°, the sides need to be perpendicular, so the slopes need to relate as follows:
[tex]m_1=-\frac{1}{m_2}[/tex]
We don't have this relation with the slopes we calculated, so the angles are not right angles. So the quadrilateral BCDE is a parallelogram.
8)
In a parallelogram, the diagonals intersect in their middle point, so the coordinates of the intersection point 'M' are the average of the starting and ending point.
So we have that:
[tex]\begin{gathered} B(-6,5),D(2,-9)\colon \\ M=(\frac{-6+2}{2},\frac{5-9}{2})=(-\frac{4}{2},-\frac{4}{2})=(-2,-2) \\ \\ C(6,2),E(-10,-6)\colon \\ M=(\frac{6-10}{2},\frac{2-6}{2})=(-\frac{4}{2},-\frac{4}{2})=(-2,-2) \end{gathered}[/tex]
So the coordinates of the intersection point are (-2, -2).
