Question

Find the relationship between x and y such that the point p(x, y) is equidistant from the points A(_5,3) and B(7,2)

Answer 1

Distance formula:

The distance between two points say A(x1,y1) and B(x2,y2) is:-

$\sqrt{(x2-x1)^2+(y2-y1)^2}$

Given:

x1=5

x2=7

y1=3

y2=2

Also:

As the lines are equidistant hence:

Ap=Bp

Hence:

$\sqrt{(x-5)^2+(y-3)^2}=\sqrt{(x-7)^2+(y-2)^2}$

Squaring both sides we get:-

$\implies (x-5)^2+(y-3)^2=(x-7)^2+(y-2)^2$

$\implies x^2+25-10x+y^2+9-6y=x^2+49-14x+y^2+4-4y$

$\implies 25-10x+9-6y=49-14x+4-4y$

$\implies 34+4x-2y-53=0$

$\implies 4x-2y=19$

The relation is 4x-2y=19

Hope it helps you.

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Answer 2

The distance between two points say A(x1,y1) and B(x2,y2) is:-

Given:

x1=5

x2=7

y1=3

y2=2

Also:

As the lines are equidistant hence:

Ap=Bp

Hence:

Squaring both sides we g