Question

The class $x$ students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of $1$m from each other. There is a triangular lawn in the plot as shown in the Fig. The students are to sow the seeds of flowering plants on the remaining area of the plot. $\left(i\right)$ Taking A as origin, find the coordinates of the vertices of the triangle. $\left(ii\right)$ What will be the coordinates of the vertices of triangle PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?

Answer 1

$\left(i\right)$ On taking A as origin, coordinates of the vertices P, Q, and R are,

From figure: P $=\left(4,6\right)$, Q $=\left(3,2\right)$, R $=\left(6,5\right)$

Now AD is the x-axis and AB is the $y-$axis.

(ii) On taking C as origin,

Coordinates of vertices P, Q, and R are $\left(12,2\right)$ , $\left(13,6\right)$ and $\left(10,3\right)$ respectively.

Now CB is the $x-$axis and CD is the $y-$axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using the formula: Area of the triangle$=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$$\begin{gathered} =\frac{1}{2}\left[4\left(2-5\right)+3\left(5-6\right)+6\left(6-2\right)\right] \\ =\frac{1}{2}\left[-12-3+24\right] \\ =\frac{9}{2}sq.units \end{gathered}$$

(ii) Area of triangle PQR in case of origin C:

Area of a triangle $=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$$\begin{gathered} =\frac{1}{2}\left[12\left(6-3\right)+13\left(3-2\right)+10\left(2-6\right)\right] \\ =\frac{1}{2}\left[36+13-40\right] \\ =\frac{9}{2}sq.units \end{gathered}$$

This implies, that the Area of triangle PQR at origin A $=$ Area of triangle PQR at origin C

Therefore, The area is the same in both cases because the triangle remains the same no matter which point is considered as the origin.