Let the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by x-axis be k : 1.

Therefore, the coordinates of the point of division, say P(x, y) is

\( \dfrac{(-4k+1)}{(k+1)}\,\dfrac{(5k-5)}{(k+1)}\)

or P(x, y)= \( \dfrac{-4k+1}{k+1},\dfrac{5k-5}{k+1}\)

We know that y-coordinate of any point on x-axis is 0.

Therefore, \( \dfrac{5k-5}{k+1}=0\)

5k = 5

or k = 1

So, *x*-axis divides the line segment in the ratio 1:1.

Now, find the coordinates of the point of division:

P (x, y) = \( \dfrac{(-4(1)+1)}{(1+1) },\dfrac{(5(1)-5)}{(1+1)}=\dfrac{-3}{2},0\)

Answered by Pragya Singh | 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.