Question

Write the equation for the perpendicular bisector of the given line segment.

Answer 1

Answer:

$y=\frac{1}{3}x+\frac{5}{3}$

Step-by-step explanation:

Let

A(-5,5),B(-2,-4) ----> the given segment

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{-4-5}{-2+5}$

$m=\frac{-9}{3}$

$m=-3$

step 2

we know that      

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

so

the slope of the perpendicular bisector is equal to

$m_1*m_2=-1$

we have

$m_1=-3$

substitute

$(-3)*m_2=-1$

$m_2=\frac{1}{3}$

step 3

Find the midpoint segment AB

A(-5,5),B(-2,-4)

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2} ,\frac{y1+y2}{2})$

substitute the values

$M(\frac{-5-2}{2} ,\frac{5-4}{2})$

$M(-\frac{7}{2} ,\frac{1}{2})$

step 4

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{1}{3}$

$M(-\frac{7}{2} ,\frac{1}{2})$

substitute

$y-\frac{1}{2}=\frac{1}{3}(x+\frac{7}{2})$

step 5

Convert to slope intercept form

$y=mx+b$

isolate the variable y

$y-\frac{1}{2}=\frac{1}{3}x+\frac{7}{6}$

$y=\frac{1}{3}x+\frac{7}{6}+\frac{1}{2}$

$y=\frac{1}{3}x+\frac{10}{6}$

simplify

$y=\frac{1}{3}x+\frac{5}{3}$

see the attached figure to better understand the problem