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Question Number 8243 by lepan last updated on 04/Oct/16

Find the equation of the perpendicular bisector of the line joining the points (−5,4) to the point (9,−3)

$${Find}\:{the}\:{equation}\:{of}\:{the}\:{perpendicular}\:{bisector}\:{of}\:{the}\:{line}\:{joining}\:{the}\:{points}\:\left(−\mathrm{5},\mathrm{4}\right)\:{to}\:{the}\:{point}\:\left(\mathrm{9},−\mathrm{3}\right) \\ $$$$ \\ $$

Answered by sandy_suhendra last updated on 04/Oct/16

let A(−5,4) and B(9,−3) P is midpoint of AB so P(((−5+9)/2) , ((4−3)/2) )=P(2,(1/2)) the gradient of AB = m_(AB) = ((y_A −y_B )/(x_A −x_B )) = ((4+3)/(−5−9)) = − (1/2) the gradient of the perpendicular bisector = m_1 so m_(AB) × m_1 = −1 − (1/2) × m_1 = − 1 m_1 = 2 the equation of the perpendicular bisector : y − y_P = m_1 (x − x_P ) y − 2 = 2 (x − (1/2) ) y = 2x + 1

$$\mathrm{let}\:\mathrm{A}\left(−\mathrm{5},\mathrm{4}\right)\:\mathrm{and}\:\mathrm{B}\left(\mathrm{9},−\mathrm{3}\right) \\ $$$$\mathrm{P}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{so}\:\mathrm{P}\left(\frac{−\mathrm{5}+\mathrm{9}}{\mathrm{2}}\:,\:\frac{\mathrm{4}−\mathrm{3}}{\mathrm{2}}\:\right)=\mathrm{P}\left(\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{AB}\:=\:\mathrm{m}_{\mathrm{AB}} =\:\frac{\mathrm{y}_{\mathrm{A}} −\mathrm{y}_{\mathrm{B}} }{\mathrm{x}_{\mathrm{A}} −\mathrm{x}_{\mathrm{B}} }\:=\:\frac{\mathrm{4}+\mathrm{3}}{−\mathrm{5}−\mathrm{9}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:=\:\mathrm{m}_{\mathrm{1}} \\ $$$$\mathrm{so}\:\:\:\:\mathrm{m}_{\mathrm{AB}} \:×\:\mathrm{m}_{\mathrm{1}} \:=\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\mathrm{m}_{\mathrm{1}} \:=\:−\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{m}_{\mathrm{1}} \:=\:\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:−\:\mathrm{y}_{\mathrm{P}} \:=\:\mathrm{m}_{\mathrm{1}} \left(\mathrm{x}\:−\:\mathrm{x}_{\mathrm{P}} \:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:−\:\mathrm{2}\:\:\:\:=\:\mathrm{2}\:\left(\mathrm{x}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\mathrm{2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{1} \\ $$

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