Answer: y=-1.5x-1
Step-by-step explanation:
A(-8,-2) B(4,6) C(x,y) - the midpoint of the line
1. Calculate the coordinates of the midpoint of the segment using the formula:
[tex]\displaystyle\\C(x,y)=(\frac{x_A+x_B}{2} ,\frac{y_A+y_B}{2})\\\\ C(x,y)=(\frac{-8+4}{2},\frac{-2+6}{2})\\\\ C(x,y)=(\frac{-4}{2} ,\frac{4}{2})\\ C(x,y)=(-2,2)\\Thus,\ C(-2,2)[/tex]
2. Find the slope of the line AB using the formula:
[tex]\displaystyle\\The\ slope=\frac{y_B-y_A}{x_B-x_A} \\\\The\ slope=\frac{6-(-2)}{4-(-8)} \\\\The\ slope=\frac{6+2}{4+8} \\\\The\ slope=\frac{8}{12} \\\\The\ slope=\frac{2}{3}[/tex]
3. Find the slope of the line perpendicular to the line AB:
[tex]\displaystyle\\The \ \perp slope=-\frac{1}{the\ slope} \\\\The \ \perp slope=-\frac{1}{\frac{2}{3} } \\\\The \ \perp slope=-\frac{3}{2}[/tex]
4. Find an equation for the perpendicular bisector of the line segment:
[tex]\displaystyle\\The \perp\ slope=\frac{y-y_C}{x-x_C} \\\\-\frac{3}{2}=\frac{y-2}{x-(-2)} \\-\frac{3}{2} =\frac{y-2}{x+2} \\[/tex]
Multiply both parts of the equation by (x+2):
[tex]\displaystyle\\y-2=(-\frac{3}{2} )(x+2)\\\\y-2=-\frac{3}{2}x+(-\frac{3}{2})(2)} \\\\y-2=-\frac{3}{2}x-3 \\\\y-2+2=-\frac{3}{2}x-3+2\\\\ y=-\frac{3}{2}x-1\\\\ y=-1.5x-1[/tex]
