The equation to the locus of the points, equidistant from the points (2, 3) and
(-2,5) is
1) 2x -y +4 = 0
2) x2 - y² = 0
3) y² = 25
4) x=0
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Equation to the locus of a point equidistant from the points A(1,3) and B(−2,1) is
Medium
Solution
verified
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Correct option is B)
Let the point be P(x,y)
Distance between P(x,y) and A(1,3)=
(1−x)
2
+(3−y)
2
=
1+x
2
−2x+9+y
2
−6y
=
x
2
+y
2
−2x−6y+10
Distance between (x,y) and (−2,1)=
(−2−x)
2
+(1−y)
2
=
4+x
2
+4x+1+y
2
−2y
=
x
2
+y
2
+4x−2y+5
As the point (x,y) is equidistant from the two points, both the distances calculated are equal.
⇒
x
2
+y
2
−2x−6y+10
=
x
2
+y
2
+4x−2y+5
⇒x
2
+y
2
−2x−6y+10=x
2
+y
2
+4x−2y+5
⇒6x+4y=5
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