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.
\[\therefore Area of ∆ ABC = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| 2\left( 7 + 1 \right) + 5\left( - 1 - 3 \right) - 3\left( 3 - 7 \right) \right|\]
\[ = \frac{1}{2}\left| 16 - 20 + 12 \right|\]
\[ = 4\]
Since the origin is shifted to the point (−1, 3), the vertices of the ∆ABC will be \[A' \left( 2 + 1, 3 - 3 \right), B' \left( 5 + 1, 7 - 3 \right),\text{ and }C' \left( - 3 + 1, - 1 - 3 \right)\]
\[\text{ or }A' \left( 3, 0 \right), B' \left( 6, 4 \right),\text{ and }C' \left( - 2, - 4 \right)\]
Now, area of ∆A'B'C' :
\[\frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| 3\left( 4 + 4 \right) + 6\left( - 4 - 0 \right) - 2\left( 0 - 4 \right) \right|\]
\[ = 4\]

Hence, area of the triangle remains invariant.