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 $\Delta_1=\frac{1}{2} \begin{vmatrix}x & y & 1 \\ 1 & 1 & 1 \\ -4 & 3 & 1\end{vmatrix}$
$\& \Delta_2=\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1\end{vmatrix}$
Given $\frac{\Delta_1}{\Delta_2}=\frac{4}{7} \Rightarrow \frac{-2 x-5 y+7}{36}=\frac{4}{7}$ $\Rightarrow 14 x+35 y=-95 \ldots .(1)$
Equation of $BC$ is $4 x + y =-13$....(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 $AQR =\frac{1}{2} \times 1 \times 1=\frac{1}{2}$