Suppose the point P  has Cartesian coordinates $\left(x_P \, y_P\right)$ and l is a horizontal line with the equation $y \= y_l$ . We can do this without loss of generality by simply rotating the coordinate system appropriately. Now choose any point in the plane with the general coordinates $\left(x\, y\right)$.

The distance from P to $\left(x\, y\right)$ is given by $d_P\=\sqrt{{\left(x-x_P\right)}^2\+{\left(y-y_P\right)}^2}$.

The distance from l to $\left(x\, y\right)$ is given by $d_l\=\sqrt{{\left(x-x_{}\right)}^2\+{\left(y-y_l\right)}^2}\=\sqrt{{\left(y-y_l\right)}^2}$.

So, equating these distances and solving for y, we find:

$\sqrt{{\left(y-y_l\right)}^2} \= \sqrt{{\left(x-x_P\right)}^2\+{\left(y-y_P\right)}^2}$

${\left(y-y_l\right)}^2 \= {\left(x-x_P\right)}^2\+{\left(y-y_P\right)}^2$

$y^2 - \left(2\cdot y_l\right)\cdot y \+ y_l^{2 } \= x^2 - \left(2 \cdot x_P\right)\cdot x \+ x_P^2 \+ y^2 - \left(2\cdot y_P\right)\cdot y \+ y_P^2$$$

$-\left(2\cdot y_l\right)\cdot y \+ \left(2\cdot y_P\right)\cdot y \= x^2 -\left(2 \cdot x_P\right)\cdot x \+ x_P^2 \+ y_P^{2 }- y_l^{2 }$

$y \= \frac{ x^2 -\left(2 \cdot x_P\right)\cdot x \+ \left(x_P^2 \+ y_P^{2 }- y_l^2\right)}{\left(-2\cdot y_l \+ 2\cdot y_P\right)}$

Since all of the indexed variables are constants, this is simply a quadratic equation, proving that the locus of points equidistant from P and l forms a parabola.$$