Let A 1 , 1 ,   B - 4 , 3 , C - 2 , - 5 be vertices of a triangle A B C ,   P be a point on side B…

Plotting the given value in diagram we have,

Now let $P\left(x,y\right)$ be a point on $BC$, so area formed by $△ABP$ will be,

 ${\Delta }_1=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 1& 1& 1\\ -4& 3& 1\end{array}\right|=\frac{1}{2}\left(-2x-5y+7\right)$

And area of triangle formed by $△ABC$ will be, ${\Delta }_2=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -4& 3& 1\\ -2& -5& 1\end{array}\right|=\frac{36}{2}$

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

$\Rightarrow 14x+35y=-95 \cdots \left(1\right)$

Equation of $BC$ by two point form will be $4x+y=-13 \cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$ we get,

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

Now equation of $AP$ will be $y-1=\frac{1+{\displaystyle \frac{11}{7}}}{1+{\displaystyle \frac{20}{7}}}\left(x-1\right)$

$\Rightarrow 3y-2x=1$,

So $x-$intercept will be point $Q\left(\frac{-1}{2},0\right)$ and similarly from $AC$ we get $R\left(\frac{1}{2},0\right)$

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