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Misc-II · Q116

Q.The vertices of a triangle are A(1,4), B(2,3) and C(1,6). Find the equations of the perpendicular bisectors of sides of ∆ABC.

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A(1,4)A(1,4), B(2,3)B(2,3), C(1,6)C(1,6). For AB: midpoint (1.5,3.5)(1.5,3.5), slope AB =−1=-1, perpendicular slope =1=1: y−3.5=1(x−1.5)⇒x−y+2=0y-3.5=1(x-1.5) \Rightarrow x-y+2=0. For BC: midpoint (1.5,4.5)(1.5,4.5), slope BC =−3=-3, perpendicular slope =1/3=1/3: y−4.5=13(x−1.5)⇒3y−13.5=x−1.5⇒x−3y+12=0y-4.5=\dfrac13(x-1.5) \Rightarrow 3y-13.5=x-1.5 \Rightarrow x-3y+12=0. For CA: CA is vertical (x=1x=1) …

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