InterviewSolution
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InterviewSolution
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Let `k`be an integer such that the triangle with vertices`(k ,-3k),(5, k)`and `(-k ,2)`has area `28s qdot`units. Then the orthocentre of this triangle is at the point :`(1,-3/4)`(2) `(2,1/2)`(3) `(2,-1/2)`(4) `(1,3/4)`A. (2,-1/2)B. (1,3/4)C. (1,-3/4)D. (2,1/2) |
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Answer» Correct Answer - D Let ABC be the triangle the coordinates of whose vertices are A(h,3-k), B(5,k) and C(-k,2). It is given that Area of `DeltaABC=28` sq. units `rArr(1)/(2)|{:(k,-3k,1),(5,k,1),(-k,2,1):}|=+-28` `rArr(1)/(2)|{:(k,-3k,1),(5-k,5k,0),(-2k,2+3k,0):}|=+-28` `rArr (5-k)(2+3k)+8k^(2)=+-56` `rArr 5k^(2)+13+66=0 or 5 k^(2)+13 k` `rArr 5k^(2)+13k-46=0` `rArr (k-2)(5+3k)=0 rArr k=2` Hence, the coordinates of vertices are A(2,-6), B(5,2) and C(-2,2) . The equation of altitudes through vertices A and C are x=2 adn `3x+8y-10=0` respectively. These two altitudes instersect at (2,1/2). Hence, the coordinates of the orthocentre are (2,1/2) |
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