Question

1. a) verify that the altitude from vertex j bisects side kl in the triangle with vertices j(-5, 4), k(1, 8), and l(-1, -2). b) classify △jkl. explain your reasoning

Explanation:

Step1: Find the mid - point of side KL

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $K(1,8)$ and $L(-1,-2)$, the mid - point $M$ of $KL$ is $(\frac{1+( - 1)}{2},\frac{8+( - 2)}{2})=(0,3)$.

Step2: Find the slope of side KL

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $K(1,8)$ and $L(-1,-2)$, the slope of $KL$ is $m_{KL}=\frac{-2 - 8}{-1 - 1}=\frac{-10}{-2}=5$.

Step3: Find the slope of the altitude from J

The altitude from $J(-5,4)$ to $KL$ is perpendicular to $KL$. If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1m_2=-1$. Since $m_{KL} = 5$, the slope of the altitude $m_{altitude}=-\frac{1}{5}$.

Step4: Find the slope of the line passing through J and M

The slope of the line passing through $J(-5,4)$ and $M(0,3)$ is $m_{JM}=\frac{3 - 4}{0-( - 5)}=-\frac{1}{5}$. Since the slope of the line passing through $J$ and the mid - point of $KL$ is the same as the slope of the altitude from $J$ to $KL$, the altitude from vertex $J$ bisects side $KL$.

Step5: Classify the triangle

To classify $\triangle{JKL}$, find the lengths of the sides.
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $JK$: $d_{JK}=\sqrt{(1-( - 5))^2+(8 - 4)^2}=\sqrt{6^2+4^2}=\sqrt{36 + 16}=\sqrt{52}=2\sqrt{13}$.
For $JL$: $d_{JL}=\sqrt{(-1-( - 5))^2+(-2 - 4)^2}=\sqrt{4^2+( - 6)^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}$.
For $KL$: $d_{KL}=\sqrt{(-1 - 1)^2+(-2 - 8)^2}=\sqrt{(-2)^2+( - 10)^2}=\sqrt{4 + 100}=\sqrt{104}=2\sqrt{26}$.
Since $JK = JL$, $\triangle{JKL}$ is an isosceles triangle.

Answer:

a) The altitude from vertex J bisects side KL.
b) $\triangle{JKL}$ is an isosceles triangle.