Question

Verify that the area of the triangle with vertices (2, 3), (5, 7) and (− 3 − 1) remains invariant under the translation of axes when the origin is shifted to the point (−1, 3).

Answer 1

Let A(2, 3), B(5, 7) and C(− 3 − 1) represent the vertices of the triangle.

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|2\left(7+1\right)+5\left(-1-3\right)-3\left(3-7\right)\right| \\ =\frac{1}{2}\left|16-20+12\right| \\ =4 \end{gathered}$$

Since the origin is shifted to the point (−1, 3), the vertices of the $∆$ABC will be

$$\begin{gathered} A^{\text{'}}\left(2+1,3-3\right),B^{\text{'}}\left(5+1,7-3\right),\mathrm{and}{C}^{\text{'}}\left(-3+1,-1-3\right) \\ \mathrm{or}{A}^{\text{'}}\left(3,0\right),B^{\text{'}}\left(6,4\right),\mathrm{and}{C}^{\text{'}}\left(-2,-4\right) \end{gathered}$$

Now, area of $∆$A'B'C' :

$$\begin{gathered} \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|3\left(4+4\right)+6\left(-4-0\right)-2\left(0-4\right)\right| \\ =4 \end{gathered}$$

Hence, area of the triangle remains invariant.