Let k be in integer such that triangle with vertices `(k,-3k),(5,k)` and `(-k,2)` has area 28 sq. units. Then the orthocentre of this traingle is at the point.

Options

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

Explanation

Given, vertices of triangle are `(k,-3k),(5,k)` and `(-k,2)`.
`:. (1)/(2)|{:(k,-3k,1),(5,k,1),(-k,2,1):}|=+-28`
`implies|{:(k,-3k,1),(5,k,1),(-k,2,1):}|=+-56`
`implies k(k-2)+3k(5+k)+1(10+k^(2))=+-56`
`impliesk^(2)-2k+15k+3k^(2)+10+k^(2)=+-56`
`implies5k^(2)+13k+10=+-56`
`implies5k^(2)+13k-66=0`
or `5k^(2)+13k-46=0`
`impliesk=2` [ `:' k in I`]
Thus, the corrdinates of vertices of triangle are `A(2,-6)`, `B(5,2)` and `C(-2,2)`
Now, equation of altitude from vertex `A` is
`y-(-6)=(-1)/(((2-2)/(-2-5)))(x-2)impliesx=2`.......`(i)`

Equation of altitude from vertex `C` is
`y-2=(-1)/([(2-(-6))/(5-2)])[x-(-2)]`
`implies3x+8y-10=0`.........`(ii)`
On solving Eqs. `(i)` and `(ii)` , we get `x=2` and `y=(1)/(2)`.
`:.` Orthocentre `=(2,(1)/(2))`