Question

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

• A. (A) 6x−4y=5
• B. (B) 6x+4y=5
• C. (C) 6x+4y=7
• D. (D) 6x−4y=7

Answer 1

Answer: (B)

(B) 6x+4y=5

Answer 2

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(h−1)2+(k−3)2=(h+2)2+(k−1)2..(i)Squaring both side in equation (i) we get,⇒(h−1)2+(k−3)2=(h+2)2+(k−1)2(∵(a±b)2=a2+b2±2ab)⇒h2−2h+1+k2−6k+9=h2+4h+4+k2−2k+1⇒−2h−6k+10=4h−2k+5⇒6h+4k=5∴The locus of (h,k) is 6x+4y=5.Hence, the correct option is (B).