Answer:
![\textsf{Slope-intercept form}: \quad y=(7)/(5)x-(19)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/9h00bjb704mrw715m3gkkimuh632bjhdmn.png)
![\textsf{Standard form}: \quad 7x-5y=19](https://img.qammunity.org/2023/formulas/mathematics/college/5dpdlcuayc5hlvnj9zan8m7nzeagqul4zi.png)
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
![\textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(-6-4)/(9-(-5))=(-10)/(14)=-(5)/(7)](https://img.qammunity.org/2023/formulas/mathematics/college/pb9ioo9457lep87pzgckrljms3fg9qc3gi.png)
Midpoint of AB
![\textsf{Midpoint}=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)=\left((9+(-5))/(2),(-6+4)/(2)\right)=(2,-1)](https://img.qammunity.org/2023/formulas/mathematics/college/g0ppg06r0l4uyncrro81labd26jnl3nuzk.png)
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:
![\implies y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/college/3o566a3lb1rp5vrz6uu9ltui840a7pirsx.png)
![\implies y-(-1)=(7)/(5)(x-2)](https://img.qammunity.org/2023/formulas/mathematics/college/jxghm03vkpn0jqzdh4uk1uyfgq7f2k4ff4.png)
![\implies y+1=(7)/(5)x-(14)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/mjobchbbtgn2ri1caa6o560xu19fn17vgc.png)
![\implies y=(7)/(5)x-(14)/(5)-1](https://img.qammunity.org/2023/formulas/mathematics/college/68cju665h8dheakbwcyy9lnk41dysnkb6m.png)
![\implies y=(7)/(5)x-(19)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/gids7fzjdggilhiolk896hjj0hhsoam3z3.png)
Therefore, the formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector of segment AB is:
![\textsf{Slope-intercept form}: \quad y=(7)/(5)x-(19)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/9h00bjb704mrw715m3gkkimuh632bjhdmn.png)
![\textsf{Standard form}: \quad 7x-5y=19](https://img.qammunity.org/2023/formulas/mathematics/college/5dpdlcuayc5hlvnj9zan8m7nzeagqul4zi.png)