I'm not sure what geometric properties you're allowed to use as yet, but here's an attempt at a purely-geometric proof (trig-free).
Let's suppose, for convenience, that the point of intersection of the two lines is not on the $x$-axis and that neither line is horizontal or vertical. Call the point of intersection of each line with the $x$-axis $A$ and $B$ and the intersection point of the two lines $C$. Call the intersection of the vertical line through $C$ with the $x$-axis $D$. Looking at $\triangle ADC$, $\frac{DC}{AD}$ is the absolute value of the slope of the line that contains $A$ and $C$; similarly, $\frac{DC}{BD}$ is the absolute value of the slope of the line that contains $B$ and $C$.
If the lines are perpendicular, than $\angle ACB$ is a right angle, so $\triangle ABC$ is a right triangle, and $CD$ is the geometric mean of $AD$ and $BD$, so $AD\cdot BD=CD^2$, from which $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$, so the product of the absolute values of the slopes is $1$. Since the slopes clearly have opposite signs, their product is $-1$.
If the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $AD\cdot BD=CD^2$. If you reflect point $C$ over the $x$-axis to $C'$, $CD=C'D$ and $AD\cdot BD=CD\cdot C'D$, so by the power of a point theorem, $A$, $B$, $C$, and $C'$ lie on a circle and since $AB$ is the perpendicular bisector of $CC'$, $AB$ is a diameter of the circle, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.

Alternately, if the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $\frac{DC}{AD}=\frac{BD}{DC}$ and since $\angle ADC$ and $\angle BDC$ are both right angles, $\triangle ADC\sim\triangle CDB$, so $\angle DAC\cong\angle DCB$ and $\angle DCA\cong\angle DBC$. Now, looking at the measures of the interior angles of $\triangle ABC$, their sum must be $180°$, but $\angle ACB$ is the sum of two angles that are congruent to $\angle ABC$ and $\angle BAC$, so the measure of $\angle ACB$ must be half of $180°$, which is $90°$, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.