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 .The students are to sow the seeds of flowering plants on the remaining area of the plot.

**(i)** Taking A as origin, find the coordinates of the vertices of the triangle.

**(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?

Asked by Pragya Singh | 1 year ago | 113

**(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

= \( \dfrac{1}{2}\) × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \( \dfrac{1}{2}\) [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= \( \dfrac{1}{2}\) (- 12 – 3 + 24 )

= \( \dfrac{9}{2}\) sq unit

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

Area of a triangle = \( \dfrac{1}{2}\) × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \( \dfrac{1}{2}\) [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= \( \dfrac{1}{2}\) ( 36 + 13 – 40)

= \( \dfrac{9}{2}\) 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 origin.

Answered by Abhisek | 1 year agoIn the determine whether the given quadratic equations have real roots and if so, And the roots 3x^{2} – 2x + 2 = 0

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Answer the following questions:-

**(i) **Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

**(ii)** Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

**(iii)** Name the type of triangle PQR formed by the point \( P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )\).

Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.