Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (-9,3) and (-3,-1)

Answer 1

`3x-2y+20=0`

Explanation:

`\mathrm{Slope\ of\ the\ line\ passing\ through\ (-9,3)\ and\ (-3,-1)\ is:}\\\mathrm{m=(-1-3)/(-3-(-9))=(-4)/(-3+9)=(-4)/(6)=-(2)/(3)}\\\\`

`\mathrm{Midpoint\ of\ the\ line\ segment\ with\ endpoints\ (-9,3)\ and\ (-3,-1)\ is:}\\\mathrm{((-9-3)/(2),(3-1)/(2))=(-6,1)}\\\mathrm{Therefore,\ the\ perpendicular\ bisector\ passes\ through\ (6,-1).\ Let\ the\ slope\ of\ the}\\\mathrm{bisector\ be\ m_1.}\\\mathrm{From\ the\ condition\ of\ perpendicular\ lines,}\\\mathrm{m.m_1=-1}\\\mathrm{or,\ (-2)/(3).m_1=-1}\\\mathrm{or,\ m_1=(3)/(2)}\\`

`\mathrm{Equation\ of\ the\ line\ having\ slope\ (3)/(2)\ and \ passing\ through\ (-6,1)\ is:}\\\mathrm{y-1=(3)/(2)(x+6)}\\\mathrm{or,\ 2y-2=3x+18}\\\mathrm{or,\ 3x-2y+20=0\ is\ the\ required\ equation\ of\ the\ perpendicular\ bisector.}`