Answer:
[tex]\textsf{Slope-intercept form}: \quad y=\dfrac{7}{5}x-\dfrac{19}{5}[/tex]
[tex]\textsf{Standard form}: \quad 7x-5y=19[/tex]
Step-by-step explanation:
A perpendicular bisector is a line that intersects another line segment at 90°, dividing it into two equal parts.
To find the perpendicular bisector of segment AB, find the slope of AB and the midpoint of AB.
Define the points:
- Let (x₁, y₁) = A(-5, 4)
- Let (x₂, y₂) = B(9, -6)
Slope of AB
[tex]\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-6-4}{9-(-5)}=\dfrac{-10}{14}=-\dfrac{5}{7}[/tex]
Midpoint of AB
[tex]\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)=\left(\dfrac{9+(-5)}{2},\dfrac{-6+4}{2}\right)=(2,-1)[/tex]
If two lines are perpendicular to each other, their slopes are negative reciprocals.
Therefore, the slope of the line that is perpendicular to line segment AB is ⁷/₅.
Substitute the found perpendicular slope and the midpoint of AB into the point-slope formula to create an equation for the line that is the perpendicular bisector of line segment AB:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-(-1)=\dfrac{7}{5}(x-2)[/tex]
[tex]\implies y+1=\dfrac{7}{5}x-\dfrac{14}{5}[/tex]
[tex]\implies y=\dfrac{7}{5}x-\dfrac{14}{5}-1[/tex]
[tex]\implies y=\dfrac{7}{5}x-\dfrac{19}{5}[/tex]
Therefore, the formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector of segment AB is:
[tex]\textsf{Slope-intercept form}: \quad y=\dfrac{7}{5}x-\dfrac{19}{5}[/tex]
[tex]\textsf{Standard form}: \quad 7x-5y=19[/tex]
