Question

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

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

Answer 1

The correct option is D $(2,\frac{1}{2})$

Given, vertices of triangle are (k,-3k),(5,k) and (-k,2).
$\therefore \frac{1}{2}\left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right|=\pm 28$
$\Rightarrow \left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right|=\pm 56$
$\Rightarrow k(k-2)+3k(5+k)+1(10+k^2)=\pm 56\Rightarrow k^2-2k+15k+3k^2+10+k^2=\pm 56\Rightarrow 5k^2+13k+10=\pm 56\Rightarrow 5k^2+13k-66=0or5k^2+13k-46=0\Rightarrow k=2[\because kϵI]$
Thus, the coordinates of vertices of triangle are $A(2,-6),B(5,2)$ and $C(-2,2)$
Now, equation of altitude from vetex A is
$y-(-6)=\frac{-1}{\left(\frac{2-2}{-2-5}\right)}(x-2)\Rightarrow x=2$ ....(i)



Equation of altitude from vertex C is
$y-2=\frac{-1}{\left[\frac{2-(-6)}{5-2}\right]}[x-(-2\left)\right]\Rightarrow 3x+8y-10=0....\left(ii\right)$
On solving Eqs. (i) and (ii), we get $x=2$
and $y=\frac{1}{2}$
$\therefore$ Orthocentre = $(2,\frac{1}{2})$