Question

whats the equation for the perpendicular bisector of the line segment whose endpoints are (-5,3) and (3,7)?

Answer 1

Answer:

The equation of perpendicular bisector of the line segment passing through  (-5,3) and (3,7) is: $y = -2x+3$

Step-by-step explanation:

Given points are:

(-5,3) and (3,7)

The perpendicular bisector of line segment formed by given points will pass through the mid-point of the line segment.

First of all we have to find the slope and mid-point of given line

Here

(x1,y1) = (-5,3)

(x2,y2) = (3,7)

The slope will be:

$m = \frac{y_2-y_1}{x_2-x_1}\\m = \frac{7-3}{3+5}\\m = \frac{4}{8}\\m = \frac{1}{2}$

The mid-point will be:

$(x,y) = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})\\ = (\frac{-5+3}{2},\frac{3+7}{2})\\= (\frac{-2}{2},\frac{10}{2})\\=(-1,5)$

Let m1 be the slope of the perpendicular bisector

Then using, "Product of slopes of perpendicular lines is -1"

$m.m_1 = -1\\\frac{1}{2}.m_1 = -1\\m_1 = -1*2\\m_1 = -2$

We have to find the equation of a line with slope -2 and passing through (-1,5)

The slope-intercept form is given by:

$y = mx+b\\y = -2x+b$

Putting the point (-1,5) in the equation

$5 = -2(-1)+b\\5 = 2+b\\b = 5-2\\b = 3$

The final equation is:

$y = -2x+3$

Hence,

The equation of perpendicular bisector of the line segment passing through  (-5,3) and (3,7) is: $y = -2x+3$