Answer to Question #275521 in Geometry for Rohit

Question

Find the equation of line which bisects the join of (2,-5) and (6,3) and bisects the join of (-1.1) and (-5,7)

Answer 1

(a) Let us call points AB

The right bisector of a line segment bisects the line segment at 90∘

The end-points of the line segment are given as A(2,-5) and B(6,3).

$Accordingly,\ mid-point\ of\ AB = (\frac{2+6}{2},\frac{-5+3}{2})=(4,-1)$

And slope of AB$=\frac{3--5}{6-2}=2$

therefore the slope of line perpendicular to AB$=-\frac{1}{2}$

Thus equation of the line passing through (4,-1) and having a slope of $-\frac{1}{2}$ is given by,

$(y+1)=-\frac{1}{2}(x-4)$

$y+\frac{1}{2}x=1$

thus the equation of the line is $y+\frac{1}{2}x=1$

(b)Let us call points CD

The right bisector of a line segment bisects the line segment at 90∘

The end-points of the line segment are given as C(-1,1) and D(-5,7).

$Accordingly,\ mid-point\ of\ CD = (\frac{-1-5}{2},\frac{1+7}{2})=(-3,4)$

And slope of CD $=\frac{7-1}{-5--1}=\frac{-3}{2}$

therefore the slope of line perpendicular to CD$=\frac{2}{3}$

Thus equation of the line passing through (-3,4) and having a slope of $\frac{2}{3}$ is given by,

$(y-4)=\frac{2}{3}(x+3)$

$y-\frac{2}{3}x=6$

thus the equation of the line is $y-\frac{2}{3}x=6$