Let A(1,1),B(-4,3),C(-2,-5) be vertices of a triangle ABC,P be a point on side BC, and Δ _1 and Δ _2 be the areas of triangle APB and ABC.…

Plotting the given value in diagram we have,

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

$${\Delta }_{1}=\frac{1}{2}|\begin{matrix}x & y & 1 \\ 1 & 1 & 1 \\ -4 & 3 & 1\end{matrix}|=\frac{1}{2}(-2x-5y+7)$$Δ1​=21​∣x1−4​y13​111​∣=21​(−2x−5y+7)

And area of triangle formed by $\triangle ABC$△ABC will be, ${\Delta }_{2}=\frac{1}{2}|\begin{matrix}1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1\end{matrix}|=\frac{36}{2}$Δ2​=21​∣1−4−2​13−5​111​∣=236​

Given $\frac{{\Delta }_{1}}{{\Delta }_{2}}=\frac{4}{7}\Rightarrow \frac{-2x-5y+7}{36}=\frac{4}{7}$Δ2​Δ1​​=74​⇒36−2x−5y+7​=74​

$$\Rightarrow 14x+35y=-95\cdots (1)$$⇒14x+35y=−95⋯(1)

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

Solving equation $(1)$(1) and $(2)$(2) we get,

Point $P(\frac{-20}{7},\frac{-11}{7})$P(7−20​,7−11​)

Now equation of $AP$AP will be $y-1=\frac{1+\frac{11}{7}}{1+\frac{20}{7}}(x-1)$y−1=1+720​1+711​​(x−1)

$\Rightarrow 3y-2x=1$⇒3y−2x=1,

So $x-$x−intercept will be point $Q(\frac{-1}{2},0)$Q(2−1​,0) and similarly from $AC$AC we get $R(\frac{1}{2},0)$R(21​,0)

So Area of triangle $AQR=\frac{1}{2}\times 1\times 1=\frac{1}{2}$AQR=21​×1×1=21​