Question

Find the equation of the locus of a point which is equidistant from the points (1,3) and (-2,1)

Answer 1

Let (x,y)(x,y) be the unknown point. It's distance from the yy axis is given by |x||x| . It's distance from the point (−6,4)(−6,4) is given by

d=(x−(−6))2+(y−4)2−−−−−−−−−−−−−−−−−−√d=(x−(−6))2+(y−4)2

Now put |x|=d|x|=d , and you're done, though you could, of course, simplify the equation:

x2=(x+6)2+(y−4)2x2=(x+6)2+(y−4)2 and then simplify a bit more.

Answer 2

AnswEr:

Let P (h,k) be any point on the locus.

Then,

$\qquad \sf \: PA = PB \\ \\ \\ \implies \sf {PA}^{2} = {PB}^{2} \\ \\ \\ \implies \tt \: {(h - 1)}^{2} + {(k - 3)}^{2} \\ \\ \\ \implies \tt \: {(h + 2)}^{2} + {(k - 1)}^{2} \\ \\ \\ \implies \tt \: 6h + 4k \\ \\ \\ \implies \tt \blue {5}$

Hence, locus of ( h, k ) is 6 x + 4y = 5.