How can I find the locus of point $P (x,y)$ that moves so that it is equidistant from lines $y = x+3$ and $y = x+7$?
I take any point on the first line to be $M (x,x+3)$ and second line to be $S (x,x+7)$.
When I equate PS=PM using the distance formula, the answer comes out to be $0=5$ which make no sense!
You need to be careful with your variables.
The line $L_1$ is described by the equation $y=x+3$ and the line $L_2$ is described by $y=x+7$.
However, when trying to find the set of $(x,y)$ that are equidistant from $L_1,L_2$, you can't just use the same $x,y$ for the point and the corresponding point on the two lines.
There are at least two ways of answering this question.
First the easy way. Draw the two lines and guess what the locus will be.
From the picture, it is obvious that the answer is the set of points $(x,y)$ that satisfy $y=x+5$.
Now a harder way.
First figure out the (minimum) distance from $(x,y)$ to $L_1$. One way to find this point is to minimize the distance (squared), that is minimizer $(x-t)^2+(y-(t+3))^2$ over $t$. Differentiating with respect to $t$ and setting to zero gives $t = \frac{y+x-3}{2}$, and substituting this into the distance formula gives the minimum distance as $\frac{|y+x-3|}{2}$.
Repeating this process for $L_2$ gives the distance from $(x,y)$ to $L_2$ as $\frac{|y+x-7|}{2}$.
Equating these will give the desired result, that is, solve $|y+x-7|= |y+x-3|$. It is clear that this is equivalent to solving $|s-3|=|s-7|$ and letting $x+y=s$. Solving $|s-3|=|s-7|$ gives $s=5$, and hence the solution is $x+y = 5$.
Letting $P = (x,y)$, $M = (x,x+3)$ and $S = (x,x+7)$ and using $PS = PM$, we get $\displaystyle \sqrt{(x-x)^2 + (x+3-y)^2} = \sqrt{(x-x)^2 + (x+7-y)^2}$. This simplifies to $\displaystyle x+3-y = \pm (x+7-y)$. Because the $+$ case obviously makes no sense, we get that $x + 3 - y = -x - 7 + y$, or $2y = 2x + 10$, or $y = x + 5$.
