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) `(1,-3/4)` (2) `(2,1/2)` (3) `(2,-1/2)` (4) `(1,3/4)`

Options

A. (2,-1/2)
B. (1,3/4)
C. (1,-3/4)
D. (2,1/2)

Answer

D

Explanation

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)