Question

The co-ordinates of the vertices A, B, C of a triangle are $(6,3),(-3,5),(4,-2)$ respectively and P is any point $(x,y),$ then the ratio of areas of triangles PBC and ABC is

• A. $\left|\begin{array}{c}x-y-2\end{array}\right|:7$
• B. $\left|\begin{array}{c}x+y+2\end{array}\right|:7$
• C. $\left|\begin{array}{c}x+y-2\end{array}\right|:7$
• D. None of these

Answer 1

The correct option is C $\left|\begin{array}{c}x+y-2\end{array}\right|:7$

Let $P=(x,y)$

We have area of $△PBC=\left|\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -3& 5& 1\\ 4& -2& 1\end{array}\right|\right|$

$=\frac{1}{2}\left|[x(5+2)-3(-2-y)+4(y-5)]\right|$

$=\frac{1}{2}|7x+7y-14|=\frac{7}{2}|x+y-2|$

Area of $△ABC=\left|\frac{1}{2}\left|\begin{array}{ccc}6& 3& 1\\ -3& 5& 1\\ 4& -2& 1\end{array}\right|\right|$

$=\frac{1}{2}\left|[6(5+2)-3(-2-3)+4(3-5)]\right|$

$=\frac{1}{2}|42+15-8|=\frac{49}{2}$

$\therefore \frac{area△PBC}{area△ABC}=\frac{|x+y-2|}{7}$