Question

The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is:

• (A) $6x-4y=5$
• (B) $6x+4y=5$
• (C) $6x+4y=7$
• (D) $6x-4y=7$

Answer 1

The Correct Option is B

Explanation:
Let $P(h,k)$ be any point on the locus which is equidistant from the point A(1, 3) and B(-2, 1).According to question,$PA=PB$$\sqrt{(h-1{)}^2+(k-3{)}^2}=\sqrt{(h+2{)}^2+(k-1{)}^2}$..(i)Squaring both side in equation (i) we get,$\Rightarrow (h-1{)}^2+(k-3{)}^2=(h+2{)}^2+(k-1{)}^2$$(\because (a\pm b{)}^2=a^2+b^2\pm 2ab)$$\Rightarrow h^2-2h+1+k^2-6k+9=h^2+4h+4+k^2-2k+1$$\Rightarrow -2h-6k+10=4h-2k+5$$\Rightarrow 6h+4k=5$$\therefore$The locus of $(h,k)$ is $6x+4y=5$.Hence, the correct option is (B).