Question

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

• A. $(1,\frac{3}{4})$
• B. $(1,-\frac{3}{4})$
• C. $(2,\frac{1}{2})$
• D. $(2,-\frac{1}{2})$

Answer 1

Answer: (C)

$(2,\frac{1}{2})$

Given vertices of triangle are $(k,-3k),(5,-k)and(-k,2)$

Area of triangle =28

$\to \frac{1}{2}|k(k-2)-5(-2-3k)-k(-3k-k)|=28$

$\to k^2-2k+10+15k+4k^2=\pm 56\to 5k^2+13k-46=0and\to 5k^2+13k+66=0$

$\to k=2$

$\therefore$ the cordinates of vertices of triangle are A(2,-6),B(5,2) and C(-2,2)

Now ,Equation of altitude from vertex A,

$y-(-6)=-\frac{1}{\frac{2-2}{2-5}}(x-2)$

$\to x=2-\left(1\right)$

Now equation of alitude from vertex c is

$y-2=-\frac{1}{\frac{-6-2}{2-5}}(x-(-2\left)\right)$

$\to 3x+8y-10=0$$-\left(2\right)$

On solving 1 and 2 we get x=2 and y=$\frac{1}{2}$

Therefore Orhocenter is $(2,\frac{1}{2})$