" The point equidistant to the lines "4x+3y+10=0,5x-12y+26=0,7x+24y-50=0" is "

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

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

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

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

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

A point equidistant from the line 4x+3y+10=0,\ 5x-12 y+26=0\ a n d\ 7x+24 y-50=0 is a. (1,-1) b. (1,1) c. (0,0) d. (0,1)

Show that the origin is equidistant from the lines 4x+3y+10=0;5x-12y+26=0 and 7x+24y=50

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