A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is

Given equations are:

4x + 3y + 10 = 0 …(i)

5x – 12y + 26 = 0 …(ii)

and 7x + 24y – 50 = 0 …(iii)

Let (p, q) be the point which is equidistant from the given lines.

Now, we find the distance of (p, q) from the given lines.

We know that,

the distance d of a point P(x₀, y₀) from the line Ax + By + C = 0 is given by

Distance of (p, q) from eq. (i) is

Distance of (p, q) from eq. (ii) is

Distance of (p, q) from eq. (iii) is

Given that (p, q) is equidistant from the given lines

Let put the value of (p, q) = (0, 0) in above equation, we get

Hence, the correct option is (c)