Question

A triangle $ABC$ is formed by three lines $x+y+2=0,x-2y+5=0$ and $7x+y-10=0$. $P$ is a point inside the triangle $ABC$ such that areas of the triangles $PAB,PBC$ and $PCA$ are equal. If the co-ordinates of the point $P$ are $(a,b)$ and the area of the triangle ABC is $\delta$, then find $\left(a+b+\delta \right).$

Answer 1

Given that the lines $x+y+2=0,x-2y+5=0$ and $7x+y-10=0$ forms a $△ABC$, so the intersection points of these three lines are the coordinates of the vertices of $△ABC$.

Intersection of $x+y+2=0,x-2y+5=0$ is $x=-3,y=1$

Intersection of $x+y+2=0,7x+y-10=0=0$ is $x=2,y=-4$

Intersection of $x-2y+5=0,7x+y-10=0=0$ is $x=1,y=3$

$P$ is the point inside the $△ABC$.

Since area of $△PAB,△PBC$ and $△PCA$ are equal.

Therefore, $P$ is the centroid of $△ABC$.

$\therefore (a,b)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$\Rightarrow a=0\text{and}b=0$

Now area of $△ABC$-

$ar\left(△ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}-3& 1& 1\\ 2& -4& 1\\ 1& 3& 1\end{array}\right|=15$

$\therefore \delta =15$

$\therefore a+b+\delta =0+0+15=15$

Hence the correct answer is $15$.