If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

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})] = 0

[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0

2x – y + 7y – 14 = 0

2x + 6y – 14 = 0

x + 3y – 7 = 0.

Which is the required result.

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.