Answer:
The equation of perpendicular bisector of the line segment passing through (-5,3) and (3,7) is: [tex]y = -2x+3[/tex]
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:
[tex]m = \frac{y_2-y_1}{x_2-x_1}\\m = \frac{7-3}{3+5}\\m = \frac{4}{8}\\m = \frac{1}{2}[/tex]
The mid-point will be:
[tex](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)[/tex]
Let m1 be the slope of the perpendicular bisector
Then using, "Product of slopes of perpendicular lines is -1"
[tex]m.m_1 = -1\\\frac{1}{2}.m_1 = -1\\m_1 = -1*2\\m_1 = -2[/tex]
We have to find the equation of a line with slope -2 and passing through (-1,5)
The slope-intercept form is given by:
[tex]y = mx+b\\y = -2x+b[/tex]
Putting the point (-1,5) in the equation
[tex]5 = -2(-1)+b\\5 = 2+b\\b = 5-2\\b = 3[/tex]
The final equation is:
[tex]y = -2x+3[/tex]
Hence,
The equation of perpendicular bisector of the line segment passing through (-5,3) and (3,7) is: [tex]y = -2x+3[/tex]
