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).
\[\therefore\text{ Area of triangle }ABC = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ \Rightarrow\text{ Area of triangle }ABC = \frac{1}{2}\left| 4\left( 10 + 2 \right) + 7\left( - 2 - 6 \right) + 1\left( 6 - 10 \right) \right|\]
\[ \Rightarrow\text{ Area of triangle ABC }= \frac{1}{2}\left| 48 - 56 - 4 \right| = 6\]
As the origin is shifted to the point (−2, 1), the vertices of the triangle ABC will be
\[A' \left( 4 + 2, 6 - 1 \right), B' \left( 7 + 2, 10 - 1 \right)\text{ and }C' \left( 1 + 2, - 2 - 1 \right)\]
\[\text{ or }A' \left( 6, 5 \right), B' \left( 9, 9 \right)\text{ and }C' \left( 3, - 3 \right)\]
Now, area of triangle A'B'C':
\[\frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| 6\left( 9 + 3 \right) + 9\left( - 3 - 5 \right) + 3\left( 5 - 9 \right) \right|\]
\[ = 6\]
So, in both the cases, the area of the triangle is 6 sq. units.
Hence, area of the triangle remains invariant.