Question

Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, −2) remains invariant under the translation of axes when the origin is shifted to the point (−2, 1).

Answer 1

Let the vertices of the given triangle be A(4, 6), B(7, 10) and C(1,− 2).

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\frac{1}{2}\left|4\left(10+2\right)+7\left(-2-6\right)+1\left(6-10\right)\right| \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\frac{1}{2}\left|48-56-4\right|=6 \end{gathered}$$

As the origin is shifted to the point (−2, 1), the vertices of the triangle ABC will be

$$\begin{gathered} A^{\text{'}}\left(4+2,6-1\right),B^{\text{'}}\left(7+2,10-1\right)\mathrm{and}{C}^{\text{'}}\left(1+2,-2-1\right) \\ \mathrm{or}{A}^{\text{'}}\left(6,5\right),B^{\text{'}}\left(9,9\right)\mathrm{and}{C}^{\text{'}}\left(3,-3\right) \end{gathered}$$

Now, area of triangle A'B'C':

$$\begin{gathered} \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|6\left(9+3\right)+9\left(-3-5\right)+3\left(5-9\right)\right| \\ =6 \end{gathered}$$

So, in both the cases, the area of the triangle is 6 sq. units.

Hence, area of the triangle remains invariant.