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5.3 · Q55

Q.Find equations of altitudes of the triangle whose vertices are A(2,5), B(6,−1) and C(−4,−3).

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Vertices A(2,5)A(2,5), B(6,−1)B(6,-1), C(−4,−3)C(-4,-3). Slope BC=−3+1−4−6=−2−10=15BC=\dfrac{-3+1}{-4-6}=\dfrac{-2}{-10}=\dfrac15, so the altitude from AA has slope −5-5: y−5=−5(x−2)⇒5x+y−15=0y-5=-5(x-2) \Rightarrow 5x+y-15=0.

Slope AC=−3−5−4−2=−8−6=43AC=\dfrac{-3-5}{-4-2}=\dfrac{-8}{-6}=\dfrac43, so the altitude from BB has slope −34-\dfrac34: y+1=−34(x−6)⇒4y+4=−3x+18⇒3x+4y−14=0y+1=-\dfrac34(x-6) \Rightarrow 4y+4=-3x+18 \Rightarrow 3x+4y-14=0. …

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