Question

Let $k$ be an integer such that the triangle with vertices $(k,–3k),(5,k)\text{and}(–k,2)$ 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 the triangle are $(k,–3k),(5,k)\text{and}(–k,2)$
Now, the area of triangle is
$\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|=28\Rightarrow \frac{1}{2}\left|\begin{array}{cccc}k& 5& -k& k\\ -3k& k& 2& -3k\end{array}\right|=28\Rightarrow |(k^2+15k)+(10+k^2)+(3k^2-2k)|=56\Rightarrow |5k^2+13k+10|=56\Rightarrow 5k^2+13k+10=56(\because 5k^2+13k+10>0)\Rightarrow 5k^2+13k-46=0\Rightarrow (5k+23)(k-2)=0\Rightarrow k=2(\because k\text{is integer})$

Now vertices are$A(2,–6),B(5,2),C(–2,2)$
Equation of the altitude dropped from vertex $A$ is $x=2$
Equation of the altitude dropped from vertex $C$ is $3x+8y–10=0$
On solving the equations of altitude we get, Orthocentre $=(2,\frac{1}{2})$