Question

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

Answer 1

Step-by-step explanation:

Let the point be P(x,y)

Distance between P(x,y) and A(1,3)

=√[ (1−x)² +(3−y)²] = √[1+x² −2x+9+y²-6y]

= √[x²+y² −2x−6y+10]

Distance between (x,y) and (−2,1)

= √[(−2−x)²+(1−y)²]

= √[4+x²+4x+1+y²−2]

= √[x²+y²+4x−2y+5]

As the point (x,y) is equidistant from the two points, both the distances calculated are equal.

⇒ √[x²+y²−2x−6y+10] = √[x²+y²+4x−2y+5]

⇒x²+y²−2x−6y+10=x²+y²+4x−2y+5

⇒6x+4y=5

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