The Total solution for NCERT class 6-12
Find a relation betweenx and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Point (x, y) is equidistant from (3, 6) and (– 3, 4).
36-16 = 6x+6x+12y-8y
20 = 12x+4y
3x+y = 5
3x+y-5 = 0