What is the best solution of coordinate geometry question .Give me the best and easy way to solve this 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. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.(ii) 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?

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# 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. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin? Q.5(2)

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(ii) Taking C as origin,

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

Here 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 formula: Area of a triangle = 1/2 × [x

_{1}(y_{2}– y_{3}) + x_{2}(y_{3}– y_{1}) + x_{3}(y_{1}– y_{2})]= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ (- 12 – 3 + 24 )

= 9/2 sq unit

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

Area of a triangle = 1/2 × [x

_{1}(y_{2}– y_{3}) + x_{2}(y_{3}– y_{1}) + x_{3}(y_{1}– y_{2})]= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= ½ ( 36 + 13 – 40)

= 9/2 sq unit

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

Area is same in both case because triangle remains the same no matter which point is considered as origi