Question

Find a relation between $x$ and $y$ such that the point $P(9x,y)$ is equidistant from the points $A(7,0)$ and $B(0,5)$.

Answer 1

We have,

Point P is equidistant from point A and B.

So,

Given that,$P(9x,y)$

$A(x_1,y_1)=(7,0)$

$B(x_2,y_2)=(0,5)$

Then, according to given,

$PA=PB$

$\sqrt{{(9x-7)}^2+{(y-0)}^2}=\sqrt{{(9x-0)}^2+{(y-5)}^2}$

On squaring both side and we get,

${(9x-7)}^2+y^2=81x^2+{(y-5)}^2$

$\Rightarrow 81x^2+49-126x+y^2=81x^2+y^2+25-10y$

$\Rightarrow 49-126x=25-10y$

$\Rightarrow 10y-126x=25-49$

$\Rightarrow 126x-10y=24$

$\Rightarrow 2(63x-5y)=12$

$\Rightarrow 63x-5y=6$

$\Rightarrow 63x-5y-6=0$.