Question

If B is the midpoint of AC, with point A at  (5, 2) and point C at (9, 0), what is the equation of the line perpendicular to AC that passes through B? A. y = - ½ x – 13 B. y = 2x – 13 C. y = - ½ x + 5 D. y = 2x + 5

Answer 1

The coordinates of point A, (x1, y1)=(5, 2).

The coordinates of point C, (x2, y2)=(9, 0).

Let (xm,ym) be the coordinates of the midpoint B.

Using midpoint formula, coordinates of the midpoint B can be found as,

$\begin{gathered} (x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ =(\frac{5+9}{2},\text{ }\frac{2+0}{2}) \\ =(\frac{14}{2},\text{ 1)} \\ =(7,1) \end{gathered}$

The slope of the line AC is,

$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ =\frac{0-2}{9-5} \\ =\frac{-2}{4} \\ =\frac{-1}{2} \end{gathered}$

The slope of the line perpendicular to AC is,

$\begin{gathered} M=\frac{-1}{m} \\ =\frac{-1}{\frac{-1}{2}} \\ =2 \end{gathered}$

Now using slope-point formula, the equation of a line perpendicular to AC and passing through point B can be found as,

$\begin{gathered} M=\frac{y-y_m}{x-x_m}\text{ (Slope-point formula)} \\ 2=\frac{y-1}{x-7} \\ 2x-7\times2=y-1 \\ 2x-14=y-1 \\ 2x-13=y \end{gathered}$

Therefore, the equation of a line perpendicular to AC and passing through point B is y=2x-13.