Question

Area of the triangle formed by the points$\left((a+3)(a+4),a+3\right),\left((a+2)(a+3),(a+2)\right)\mathrm{and}\left((a+1)(a+2),(a+1)\right)$ is
(a) 25a²
(b) 5a²
(c) 24a²
(d) none of these

Answer 1

(d) none of these

The given points are $(\left\{a+3\left)\right(a+4),\left(a+3\right)\right\},\left\{(a+2)(a+3),(a+2)\right\}\mathrm{and}\left\{(a+1)(a+2),(a+1)\right\}$.

Let A be the area of the triangle formed by these points.

$$\begin{gathered} \mathrm{Then,}A=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \Rightarrow A=\frac{1}{2}\left[\left(a+3\right)\left(a+4\right)\left(a+2-a-1\right)+\left(a+2\right)\left(a+3\right)\left(a+1-a-3\right)+\left(a+1\right)\left(a+2\right)\left(a+3-a-2\right)\right] \\ \Rightarrow A=\frac{1}{2}\left[\left(a+3\right)\left(a+4\right)-2\left(a+2\right)\left(a+3\right)+\left(a+1\right)\left(a+2\right)\right] \\ \Rightarrow A=\frac{1}{2}\left[a^2+7a+12-2a^2-10a-12+a^2+3a+2\right] \\ \Rightarrow A=1 \end{gathered}$$