The correct option is B (0,0)
Given equation are
4x+3y+10=0 …(i)
5x−12y+26=0 …(ii)
7x+24y−50=0 …(iii)
Let (p,q) be the point which is equidistant from the given lines
∵ Distance of point (x1,y1) from the line ax+by+c=0 is
D=|ax1+by1+c|√a2+b2
Distance of (p,q) from equation (i) is
d=|4p+3q+10|√(42+32)=|4p+3q+10|5 …(iv)
Distance of (p,q) from equation (ii) is
d=|5p−12q+26|√(52+122)=|5p−12q+26|13 …(v)
Distance of (p,q) from equation (iii) is
d=|7p+24q−50|√(72+(24)2)=|7p+24q−50|25 …(vi)
Given that (p,q) is equidistant from the given lines, then
|4p+3q+10|5=|5p−12q+26|13=|7p+24q−50|25
By using the given options, if we put (0,0), then
|10|5=|26|13=|−50|25=2
So, (0,0) is the point which is equidistant from the given lines.
Hence, Option (C) is correct.