Determine an equation for the right bisector of the line segment with endpoints J(-5,3) and K(3,-2)

Answer:[tex]y=1.6x+2.1[/tex]Step-by-step explanation:we know thatThe right bisector of the line segment JK is a perpendicular line to the segment JK that pass through the midpoint of segment JKstep 1Find the midpoint JKThe formula to calculate the midpoint between two points is equal to[tex]M(\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]we have[tex]J(-5,3),K(3,-2)[/tex]substitute the values[tex]M(\frac{-5+3}{2},\frac{3-2}{2})[/tex][tex]M(\frac{-2}{2},\frac{1}{2})[/tex][tex]M(-1,\frac{1}{2})[/tex]step 2Find the slope JKThe formula to calculate the slope between two points is equal to [tex]m=\frac{y2-y1}{x2-x1}[/tex] we have[tex]J(-5,3),K(3,-2)[/tex]substitute[tex]m=\frac{-2-3}{3+5}[/tex] [tex]m=\frac{-5}{8}[/tex] [tex]m=-\frac{5}{8}[/tex] step 3Find the slope of the line perpendicular to the segment JKwe know thatIf two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)[tex]m_1*m_2=-1[/tex]we have[tex]m_1=-\frac{5}{8}[/tex]  ----> slope of segment JKFind m_2substitute[tex](-\frac{5}{8})*m_2=-1[/tex][tex]m_2=\frac{8}{5}[/tex]step 4Find the equation for the right bisector of the line segment JK The equation in point slope form is equal to[tex]y-y1=m(x-x1)[/tex]we have[tex]m=\frac{8}{5}[/tex][tex]point\ M(-1,\frac{1}{2})[/tex]substitute[tex]y-\frac{1}{2}=\frac{8}{5}(x+1)[/tex]Convert to slope intercept formisolate the variable y[tex]y-\frac{1}{2}=\frac{8}{5}x+\frac{8}{5}[/tex][tex]y=\frac{8}{5}x+\frac{8}{5}+\frac{1}{2}[/tex][tex]y=\frac{8}{5}x+\frac{21}{10}[/tex][tex]y=1.6x+2.1[/tex]