Question

If the area of the triangle formed by the points $\left(2a,b\right),\left(a+b,2b+a\right)\mathrm{and}\left(2\mathrm{b},2\mathrm{a}\right)$ is $2sq.\mathrm{units}$, then the area of the triangle whose vertices are $\left(a+b,a-b\right),\left(3b-a,b+3a\right)and\left(3a-b,3b-a\right)$ will be

Answer 1

Find the area of the triangle

If $\left(x_1,y_1\right)$,$\left(x_2,y_3\right)$ and $\left(x_3,y_3\right)$ are the three vertices of a triangle then the area of the triangle is

Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Area of the triangle formed by the point $\left(2a,b\right),\left(a+b,2b+a\right)\mathrm{and}\left(2\mathrm{b},2\mathrm{a}\right)$ is $2\mathrm{sq}.\mathrm{units}$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}2a& b& 1\\ a+b& 2b+a& 1\\ 2b& 2a& 1\end{array}\right|=2 \\ \Rightarrow \left|\begin{array}{ccc}2a& b& 1\\ a+b& 2b+a& 1\\ 2b& 2a& 1\end{array}\right|=4 \\ \Rightarrow 3ab-3b^2=4................\left(1\right) \end{gathered}$$

Area of the triangle whose vertices are $\left(a+b,a-b\right),\left(3b-a,b+3a\right)and\left(3a-b,3b-a\right)$ is

$$\begin{gathered} \frac{1}{2}\left|\begin{array}{ccc}a+b& a-b& 1\\ 3b-a& b+3a& 1\\ 3a-b& 3b-a& 1\end{array}\right| \\ =\frac{1}{2}\left[\left(a+b\right)\left(b+3a-3b+a\right)+\left(a-b\right)\left(3a-b-3b+a\right)+\left\{\left(3b-a\right)\left(3b-a\right)-\left(3a-b\right)\left(b+3a\right)\right\}\right] \\ =\frac{1}{2}\left[12b^2-12ab\right] \\ =6\left(b^2-ab\right) \\ =6\times \frac{4}{3}\left[\because 3ab-3b^2=4\right] \\ =8 \end{gathered}$$

Hence, the area of the triangle is $8sq.\mathrm{unit}$