Let A (1, 1), B (- 4, 3), C(- 2, - 5) be vertices of a triangle...

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} = \dfrac{1}{2}\left|\begin{cases}x & y & 1 \\ 1 & 1 & 1 \\ - 4 & 3 & 1\end{cases}\right| = \dfrac{1}{2}\left(- 2x - 5y + 7\right)

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

Given $\dfrac{{\Delta}_{1}}{{\Delta}_{2}} = \dfrac{4}{7} ⇒ \dfrac{- 2x - 5y + 7}{36} = \dfrac{4}{7}$

$⇒ 14x + 35y = - 95 ⋯ \left(1\right)$

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

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

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

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

$⇒ 3y - 2x = 1$,

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

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