Question

Indicate the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5). Enter your answer into the blank equation box. (Do not use spaces.)

Answer 1

Equation of a line is $x+3y =-3$.

What is a perpendicular bisector of the line segment?

A perpendicular bisector is a line that cuts a line segment connecting two points exactly in half at a 90 degree angle. To find the perpendicular bisector of two points, all you need to do is find their midpoint and negative reciprocal, and plug these answers into the equation for a line in slope-intercept form.

Given that,

Endpoints of the line segment are ($x_{1},y_{1}$) = (4, 1) and ($x_{2},y_{2}$) = (2, -5).

First find the midpoints of the given line segment.

M = $\left(\frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2} }{2}\Right)$

    =  $\left(\frac{4+2 }{2},\frac{1-5 }{2}\Right)$

M   =  $(3,-2)$

Now, Find the slope of the line :

It is perpendicular to the line with (4,1) and (2,-5)

Slope between ($x_{1},y_{1}$) and ($x_{2},y_{2}$) = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

so,

the slope between (4,1) and (2,-5)  =  $\frac{-5-1 }{2-4 }$

                                                         = 3

perpendicular lines have slopes the multiply to get -1

3 times m=-1

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

The equation of a line that has a slope of m and passes through the midpoints M(3,-2)  is

$y-y_{1} =m(x-x_{1} )$

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

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

if we want slope intercept form

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

$y= \frac{-1}{3} x-1$

If we want standard form

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

$x+3y =-3$

Hence, Equation of a line is $x+3y =-3$.

To learn more about perpendicular bisector of the line segment from the given link:

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