Doubts on locus and its equation "Find the equation to the locus of a point which is collinear with points $M(a,0)$ and $N(0,b)$."  
The answer is $- x/a + y/b$. How I tried to find the solution:
$P$ is a point whose assigned coordinates are $(x,y)$, therefore $\vert MN \vert =\vert  MP \vert +\vert NP \vert.$ But the solution I get by this method is incorrect.
Can someone help me find the solution?
 A: Hint.
If you draw a diagram (please do this before reading further) you will see that the gradient of the line $PM$ must be the same as the gradient of the line $PN$.  This will give you the equation you need.
A: First of all, "x/a + y/b" is just an expression and not an equation. The full answer is "$\dfrac{x}{a}+\dfrac{y}{b} = 1$".  
Second, not every point $P$ which is co-linear with $M$ and $N$ satisfies $MN = MP+NP$. Try looking at the case where $M(1,0)$, $N(0,1)$, and $P(2,-1)$. 
Finally, note that $P$ being colinear with $M$ and $N$ simply means that $M$, $N$, and $P$ lie on the same line. Since there is a unique line $\ell$ through $M$ and $N$, $P$ must lie on line $\ell$. Also, all points on line $\ell$ are colinear with $M$ and $N$. Thus, the locus is the line $\ell$, so you just need to find the equation of the line passing through $M(a,0)$ and $N(0,b)$. 
A: If you have three point now M,N and P:(x,y). The area of the triangle formed by this three points are zero(as the three are co-linear by your choice of P). Equate the expression for area to zero and you will get the equation.
A: Simply you may apply area of trangle zero with three colinear vertices (a,0), (0,b) and (x,y).
