The equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is [tex]y - 3 = \frac{-7x}{2}+ \frac{21}{4}[/tex]
Given that we have to write equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)
Let us first find the slope of given line AB
The slope "m" of the line is given as:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
Here the given points are A(-2,2) and B(5,4)
[tex]\text {Here } x_{1}=-2 ; y_{1}=2 ; x_{2}=5 ; y_{2}=4[/tex]
[tex]m=\frac{4-2}{5-(-2)}=\frac{2}{7}[/tex]
Thus the slope of line with given points is [tex]\frac{2}{7}[/tex]
We know that product of slopes of given line and slope of line perpendicular to given line is always -1
[tex]\begin{array}{l}{\text {slope of given line } \times \text { slope of perpendicular bisector }=-1} \\\\ {\frac{2}{7} \times \text { slope of perpendicular bisector }=-1} \\ \\{\text {slope of perpendicular bisector }=\frac{-7}{2}}\end{array}[/tex]
The perpendicular bisector will run through the midpoint of the given points
So let us find the midpoint of A(-2,2) and B(5,4)
The midpoint formula for given two points is given as:
[tex]\text {For two points }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right), \text { midpoint } \mathrm{m}(x, y) \text { is given as }[/tex]
[tex]m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex]
Substituting the given points A(-2,2) and B(5,4)
[tex]m(x, y)=\left(\frac{-2+5}{2}, \frac{2+4}{2}\right)=\left(\frac{3}{2}, 3\right)[/tex]
Now let us find the equation of perpendicular bisector in point slope form
The perpendicular bisector passes through points (3/2, 3) and slope -7/2
The point slope form is given as:
[tex]y - y_1 = m(x - x_1)[/tex]
[tex]\text { Substitute } \mathrm{m}=\frac{-7}{2} \text { and }\left(x_{1}, y_{1}\right)=\left(\frac{3}{2}, 3\right)[/tex]
[tex]y - 3 = \frac{-7}{2}(x - \frac{3}{2})\\\\y - 3 = \frac{-7x}{2}+ \frac{21}{4}[/tex]
Thus the equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is found out
