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

Answer

A

Explanation

We have
`(1)/(2)|{:(k,,-3k,,1),(5,,k,,1),(-k,,2,,1):}|=+-28`
`rArr5k^2+13k-46=0`
or `5k^2+13k+66=0` (no ral solution)
`rArr=K=(-23)/(5)ork=2`
Since K is an integer, `k=2`

Form the figure, slope of BC is 0. So, equation of AD is
`x=2`
Slope of AC is `1//2`.
Eqaution of line BE is `x-2y-=0`