Question

Segment CD is formed by C(-5, 9) and D(7, 5). If line t is the perpendicular bisector of segment CD , write a linear equation for t in slope-intercept form.

Answer 1

Given:

Segment CD is formed by C(-5, 9) and D(7, 5).

Line t is the perpendicular bisector of segment CD.

To find:

The linear equation for t in slope-intercept form.

Solution:

Line t is the perpendicular bisector of segment CD. It means line t passes through the midpoint of CD.

$Midpoin=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoin=\left(\dfrac{-5+7}{2},\dfrac{9+5}{2}\right)$

$Midpoin=\left(\dfrac{2}{2},\dfrac{14}{2}\right)$

$Midpoin=\left(1,7\right)$

Slope of CD is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m_{1}=\dfrac{5-9}{7-(-5)}$

$m_{1}=\dfrac{-4}{12}$

$m_{1}=\dfrac{-1}{3}$

Product of slopes of two perpendicular lines is -1.

Let slope of line t be $m_2$.

$m_1\times m_2=-1$

$\dfrac{-1}{3}\times m_2=-1$

$m_2=3$

Point slope form of a line is

$y-y_1=m(x-x_1)$

where, m is slope.

The slope of line t is 3 and it passes through (1,7). So, point slope form of line t is

$y-7=3(x-1)$

Therefore, the point slope form of line t is $y-7=3(x-1)$.