# Find a relation between x and y such that the point (x, y)

Question:

Find a relation between x and y such that the point (xy) is equidistant from the points (3, 6) and (−3, 4).

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

Let the three given points be P(x, y), A(3,6) and B(3,4).

Now let us find the distance between ‘P’ and ‘A’.

$P A=\sqrt{(x-3)^{2}+(y-6)^{2}}$

Now, let us find the distance between ‘P’ and ‘B’.

$P B=\sqrt{(x+3)^{2}+(y-4)^{2}}$

It is given that both these distances are equal. So, let us equate both the above equations,

$P A=P B$

$\sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}}$

Squaring on both sides of the equation we get,

$(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$

$x^{2}+9-6 x+y^{2}+36-12 y=x^{2}+9+6 x+y^{2}+16-8 y$

$12 x+4 y=20$

$3 x+y=5$

Hence the relationship between ' $x$ ' and ' $y$ ' based on the given condition is $3 x+y=5$.