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Question

Mathematics Question on Distance of a Point From a Line

Find the equation of the right bisector of the line segment joining the points (3,4) and (–1,2).

Answer

The right bisector of a line segment bisects the line segment at 90°.90°.
The endpoints of the line segment are given as A(3,4)A (3, 4) and B(−1,2).B (-1, 2).
Accordingly, mid-point of

AB=(3−12,4+22)=(1,3)AB = \left(\frac{3-1}{2},\frac{4+2}{2}\right)=(1,3)

Slope of AB 2−4−1−3=−2−4=12\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}

∴ Slope of the line perpendicular to AB=−1(12)=−2AB =\frac{-1}{(\frac{1}{2})}=-2
The equation of the line passing through (1,3)(1, 3) and having a slope of -2 is
(y\-3)=−2(x\-1)(y \- 3) = -2 (x \- 1)
y\-3=−2x+2y \- 3 = -2x + 2
2x+y=52x + y = 5
Thus, the required equation of the line is 2x+y=5.2x + y = 5.