Let A(1, 1), B(-4, 3) C(-2, -5) be vertices of a triangle ABC, P be a point on side BC, and ${∆}_1 \mathrm{and} {∆}_2$  be the areas of triangle APB and ABC, respectively.

If ${\mathrm{\Delta }}_1 : {\mathrm{\Delta }}_2=4 : 7$, then the area enclosed by the lines AP, AC and the x-axis is

Given ${\mathrm{\Delta }}_1=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& 1\\ 1& 1& 1\\ -4& 3& 1\end{array}\right|$ and ${\mathrm{\Delta }}_2=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -4& 3& 1\\ -2& -5& 1\end{array}\right|$

Given $\frac{{\mathrm{\Delta }}_1}{{\mathrm{\Delta }}_2}=\frac{4}{7}\Rightarrow \frac{-2\mathrm{x}-5\mathrm{y}+7}{36}=\frac{4}{7}$

$\Rightarrow 14\mathrm{x}+35\mathrm{y}=-95 ...\left(1\right)$

Equation of BC is 4x + y = -13     ….(2)

Solve equation (1) & (2)

Point  $\mathrm{P}\left(\frac{-20}{7},\frac{-11}{7}\right)$

Here point  $\mathrm{Q}\left(\frac{-1}{2},0\right) \mathrm{\&} \mathrm{R}\left(\frac{1}{2},0\right)$

So Area of triangle $\mathrm{AQR}=\frac{1}{2}\times 1\times 1=\frac{1}{2}$