Ex-7. 4, question no. 5

1.	Ex-7. 4, question no. 5
Answer» Solution:(i) Taking A as origin, coordinates of the vertices P, Q and R are,From figure: P = (4, 6), Q = (3, 2), R (6, 5)Here AD is the x-axis and AB is the y-axis.(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 == 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 == ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]= ½ ( 36 + 13 – 40)= 9/2 sq unitThis implies, Area of triangle PQR at origin A = Area of triangle PQR at origin CArea is same in both case because triangle remains the same no matter which point is considered as origin.

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