Question

Let $A(1,1),B(-4,3)C(-2,-5)$ be vertices of a triangle $ABC,P$ be a point on side $BC$, and ${\Delta }_1$ and ${\Delta }_2$ be the areas of triangle APB and ABC. Respectively.
If ${\Delta }_1:{\Delta }_2=4:7$, then the area enclosed by the lines AP, AC and the $x$-axis is

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (C)

Given ${\Delta }_1=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 1& 1& 1\\ -4& 3& 1\end{array}\right|$
$\&{\Delta }_2=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -4& 3& 1\\ -2& -5& 1\end{array}\right|$
Given $\frac{{\Delta }_1}{{\Delta }_2}=\frac{4}{7}\Rightarrow \frac{-2x-5y+7}{36}=\frac{4}{7}$ $\Rightarrow 14x+35y=-95\dots .(1)$
Equation of $BC$ is $4x+y=-13$....(2)
Solve equation (1) \& (2)
Point $P\left(\frac{-20}{7},\frac{-11}{7}\right)$
Here point $Q\left(\frac{-1}{2},0\right)\&R\left(\frac{1}{2},0\right)$
So Area of triangle $AQR=\frac{1}{2}\times 1\times 1=\frac{1}{2}$