Question

Area of the triangle formed by the points $\left(\right(a+3\left)\right(a+4)a+3),\left(\right(a+2\left)\right(a+3),(a+2\left)\right)$ and $\left(\right(a+1\left)\right(a+2\left)\right(a+1\left)\right)$

• A. $25a^2$
• B. $5a^2$
• C. $24a^2$
• D. none of these

Answer 1

The correct option is D

none of these

The given points are

$(\{a+3)(a+4),(a+3)\},\left\{\right(a+2\left)\right(a+3),(a+2\left)\right\}$ and $\left\{\right(a+1\left)\right(a+2),(a+1\left)\right\}$

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

Then,

$A=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$\Rightarrow A$

$\Rightarrow \frac{1}{2}\left[\right(a+3\left)\right(a+4\left)\right(a+2-a-1)+(a+2\left)\right(a+3\left)\right(a+1-a-3)+(a+1\left)\right(a+2\left)\right(a+3-a-2\left)\right]$

$\Rightarrow A=\frac{1}{2}\left[\right(a+3\left)\right(a+4)-2(a+2\left)\right(a+3)+(a+1\left)\right(a+2\left)\right]$

$\Rightarrow A=\frac{1}{2}[a^2+7a+12-2a^2-10a-12+a^2+3a+2]$

$\Rightarrow A=1\left[\right(a+3\left)\right(a+4\left)\right(a+2-a-1)+(a+2\left)\right(a+3\left)\right(a+1-a-3)+(a+1\left)\right(a+2\left)\right(a+3-a-2\left)\right]$

$=\frac{1}{2}\left[\right(a+3\left)\right(a+4)-2(a+2\left)\right(a+3)+(a+1\left)\right(a+2\left)\right]$

$=\frac{1}{2}[a^2+7a+12-2a^2-10a-12+a^2+3a+2]$

$=1$