If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y =0.
Let D1 be the distance between (x,y) and (2,1)
D2 be the distance between (x,y) and (1,−2)
D1=√(x−2)2+(y−1)2)
D1=√(x2−4x+4+y2−2y+1)
D1=√(x2+y2−4x−2y+5)
D2=√(x−1)2+(y+2)2)
D2=√(x2−2x+1+y2+4y+4)
D2=√(x2+y2−2x+4y+5)
Here,
D1=D2
√(x2+y2−4x−2y+5)=√(x2+y2−2x+4y+5)
−4x−2y+5=−2x+4y+5
2x+6y=0
x+3y=0
Hence proved.