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5.3 · Q56

Q.Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(−1,8), Q(4,−2) and R(−5,−3).

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Vertices P(−1,8)P(-1,8), Q(4,−2)Q(4,-2), R(−5,−3)R(-5,-3).

For side PQPQ: midpoint =(32,3)=\left(\dfrac32,3\right), slope PQ=−2−84+1=−2PQ=\dfrac{-2-8}{4+1}=-2, so perpendicular slope =12=\dfrac12. Bisector: y−3=12(x−32)⇒2y−6=x−1.5⇒2x−4y+9=0y-3=\dfrac12\left(x-\dfrac32\right) \Rightarrow 2y-6=x-1.5 \Rightarrow 2x-4y+9=0.

For side QRQR: midpoint =(−12,−52)=\left(-\dfrac12,-\dfrac52\right), slope QR=−3+2−5−4=19QR=\dfrac{-3+2}{-5-4}=\dfrac19, so perpendicular slope =−9=-9. Bisector: y+2.5=−9(x+0.5)⇒9x+y+7=0y+2.5=-9\left(x+0.5\right) \Rightarrow 9x+y+7=0. …

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