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Find equation of perpendicular bisector of the line segment joining the points (4, -3) and (3, 1).

Straight Line Eq

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Answer

Mid-point of line segment joining the points (4, -3) and (3, 1) is (4+32,−3+12)=(72,−22)=(72,−1).\Big(\dfrac{4 + 3}{2}, \dfrac{-3 + 1}{2}\Big) = \Big(\dfrac{7}{2}, \dfrac{-2}{2}\Big) = \Big(\dfrac{7}{2}, -1\Big).

Slope of line joining the points (4, -3) and (3, 1) is :

=y2−y1x2−x1=1−(−3)3−4=4−1=−4.= \dfrac{y2 - y1}{x2 - x1} \\[1em] = \dfrac{1 - (-3)}{3 - 4} \\[1em] = \dfrac{4}{-1} \\[1em] = -4.

We know that,

⇒ Product of slopes of perpendicular line = -1.

⇒ Slope of line joining the points (4, -3) and (3, 1) × Slope of perpendicular line = -1

⇒ -4 × Slope of perpendicular line = -1

⇒ Slope of perpendicular line = 14\dfrac{1}{4}.

Perpendicular bisector of the line segment joining the points (4, -3) and (3, 1) will have slope = 14\dfrac{1}{4} and will pass through the point (72,−1)\Big(\dfrac{7}{2}, -1\Big).

By point-slope form, equation of perpendicular bisector is

⇒y−y1=m(x−x1)⇒y−(−1)=14(x−72)⇒4(y+1)=x−72⇒4y+4=x−72⇒2(4y+4)=2(x−72)⇒8y+8=2x−7⇒2x−8y−7−8=0⇒2x−8y−15=0.\Rightarrow y - y1 = m(x - x1) \\[1em] \Rightarrow y - (-1) = \dfrac{1}{4}\Big(x - \dfrac{7}{2}\Big) \\[1em] \Rightarrow 4(y + 1) = x - \dfrac{7}{2} \\[1em] \Rightarrow 4y + 4 = x - \dfrac{7}{2} \\[1em] \Rightarrow 2(4y + 4) = 2\Big(x - \dfrac{7}{2}\Big) \\[1em] \Rightarrow 8y + 8 = 2x - 7 \\[1em] \Rightarrow 2x - 8y - 7 - 8 = 0 \\[1em] \Rightarrow 2x - 8y - 15 = 0.

Hence, equation of perpendicular bisector of the line segment joining the points (4, -3) and (3, 1) is 2x - 8y - 15 = 0.

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