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Misc-II · Q114

Q.The vertices of a triangle are A(1,4), B(2,3) and C(1,6). Find the equations of the 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). Slope AB=3−42−1=−1AB=\dfrac{3-4}{2-1}=-1: through A, y−4=−1(x−1)⇒x+y−5=0y-4=-1(x-1) \Rightarrow x+y-5=0. Slope BC=6−31−2=−3BC=\dfrac{6-3}{1-2}=-3: through B, y−3=−3(x−2)⇒3x+y−9=0y-3=-3(x-2) \Rightarrow 3x+y-9=0. Since AA and CC share $x=1 …

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