Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle.

Coordinates of D, E, and F are given by

D = (0+\(\dfrac{2}{2}\), -1+\( \dfrac{1}{2}\) ) = (1, 0)

E = ( 0+0/2, -1+\( \dfrac{3}{2}\)) = (0, 1)

F = ( 0+\( \dfrac{2}{2}\), 3+\( \dfrac{1}{2}\)) = (1, 2)

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})]

Area of ΔDEF =\( \dfrac{1}{2}\){1(2-1) + 1(1-0) + 0(0-2)}

= \( \dfrac{1}{2}\)(1+1) = 1

**Area of ΔDEF is 1 square units**

Area of ΔABC =\( \dfrac{1}{2}\)[0(1-3) + 2{3-(-1)} + 0(-1-1)]

= \( \dfrac{1}{2}\) {8} = 4

**Area of ΔABC is 4 square units**

Therefore, the required ratio is 1:4.

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.