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5.3 · Q57

Q.Find the co-ordinates of the orthocenter of the triangle whose vertices are A(2,−2), B(1,1) and C(−1,0).

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Vertices A(2,−2)A(2,-2), B(1,1)B(1,1), C(−1,0)C(-1,0). Altitude from A⊥BCA \perp BC: slope BC=0−1−1−1=12BC=\dfrac{0-1}{-1-1}=\dfrac12, so altitude slope =−2=-2: y+2=−2(x−2)⇒2x+y−2=0y+2=-2(x-2) \Rightarrow 2x+y-2=0 ...(i).

Altitude from B⊥ACB \perp AC: slope AC=0+2−1−2=−23AC=\dfrac{0+2}{-1-2}=-\dfrac23, so altitude slope =32=\dfrac32: y−1=32(x−1)⇒3x−2y−1=0y-1=\dfrac32(x-1) \Rightarrow 3x-2y-1=0 ...(ii). …

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