In a triangle ABC , coordinates of A are (1, 2) and the equations...

Let C\left(4, α\right) be on $x = 4$.

Now, midpoint of AC is \left(\dfrac{5}{2}, \dfrac{α + 2}{2}\right)$which lies on$x + y = 5$.</p><p>$∴ \dfrac{α + 2}{2} = \dfrac{5}{2} \to α = 3$</p><p>$∴ C\left(4, 3\right)$</p><p>Similarly, let$B\left(β\right)\left(, 5 - β\right)$be on$x + y = 5$.</p><p>The midpoint of AB$= \left(\dfrac{β + 1}{2}, \dfrac{7 - β}{2}\right)$which lies on$x = 4 \to β = 7 \to B\left(7, - 2\right)$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/a6de6614-3159-42b6-bbd3-4405f4686f51-image.png" style="width: 230px; height: 185px;" /> </p><p>$∴$Area of$Δ = \dfrac{1}{2}\left|$\begin{cases}1 & 2 & 1 \\ 7 & - 2 & 1 \\ 4 & 3 & 1\end{cases}$\right|$</p><p>$= \dfrac{1}{2}\left[1\left(- 2 - 3\right) - 2\left(7 - 4\right) + 1\left(21 + 8\right)\right] = \dfrac{1}{2}\left[18\right] = 9$ sq. units