Question

1. the endpoints of line segment ab are located at (5, -2) and (-3,10). what is the equation of the line that is perpendicular to ab and passes through the midpoint of ab? a) $y - 4=-\frac{3}{2}(x - 1)$ b) $y - 4=\frac{2}{3}(x - 1)$ c) $y - 6=\frac{3}{2}(x - 4)$ d) $y + 2=\frac{2}{3}(x - 5)$

Explanation:

Step1: Find the mid - point of line segment AB

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Given $(x_1,y_1)=(5,-2)$ and $(x_2,y_2)=(-3,10)$.
$x_m=\frac{5+( - 3)}{2}=\frac{2}{2}=1$
$y_m=\frac{-2 + 10}{2}=\frac{8}{2}=4$
The mid - point is $(1,4)$.

Step2: Find the slope of line segment AB

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$.
$m_{AB}=\frac{10-( - 2)}{-3 - 5}=\frac{10 + 2}{-8}=\frac{12}{-8}=-\frac{3}{2}$

Step3: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular line be $m_p$.
$m_{AB}\times m_p=-1$. Since $m_{AB}=-\frac{3}{2}$, then $-\frac{3}{2}\times m_p=-1$, so $m_p=\frac{2}{3}$.

Step4: Write the point - slope form of the line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Using the point $(1,4)$ and slope $m=\frac{2}{3}$, we get $y - 4=\frac{2}{3}(x - 1)$.

Answer:

B. $y - 4=\frac{2}{3}(x - 1)$