Question

the x coordinates of a point p is equidistance from a(2,-5),b(-3,6).find the coordinates of p

Answer 1

Answer:

the point which is equidistant from two points is the midpoint of the two points

Step-by-step explanation:

so the mid point of the given points is ((2-3)/2,(-5+6)/2)

=(-1/2,1/2)

Answer 2

Step-by-step explanation:

Given points are a (2,-5) b (-3,6)

Let P (x,y) be the point which is equidistant from a and b.

Here

$pa = pb \\ squaring \: on \: both \: sides \\ {pa}^{2} = {pb}^{2} \\ \: {(x - 2)}^{2} + {(y + 5)}^{2} = {(x + 3)}^{2} + {(y - 6)}^{2} \\ {x}^{2} + 4 - 4x + {y}^{2} + 25 + 10y = {x}^{2} + 9 + 6x + {y}^{2} + 36 - 12y \\ {x}^{2} + {y}^{2} - 4x + 10y + 29 = {x}^{2} + {y }^{2} +6 x - 12y + 45 \\ - 4x + 10y + 29 = 6x - 12y + 45 \\ 10x - 22y + 16 = 0 \\ 5x - 11y + 8 = 0$

Hence the equation of Locus of point p is

$5x - 11y + 8 = 0$

Hope it helped.....

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