Find the coordinates of the point $P$ on the $x$-axis such that the segments $AP$ and $BP$ make the same angle with the normal to the $x$-axis at $P$ Consider the points $A = (0, 1)$ and $B = (2, 2)$ in the plane. Find the
coordinates of the point $P$ on the $x$-axis such that the segments $AP$ and $BP$ make the same angle with the normal to the $x$-axis at $P$.
I was  trying this question but I could not get it, I was using the distance formula, and   trying to find out the normal vector to the axis at $P$. But could not able to find it out.
If anybody help me, I would be very thankful to them.
 A: Reflect the point $A$ wrt $x-$axis to get $A'(0,-1)$. The point $P$ is the intersection point of $BA'$ with $x-$axis
In your example line $BA'$ has equation $3x-2y=2$ so coordinates are $P\left(\dfrac{2}{3},\;0\right)$
A: Assume P to be ($\Delta$,0)
Now, the angles that $AP$ and $BP$ make with the x-axis are equal (as the angles that they make with the normal are equal and the normal is $\bot$ to x-axis).
Now, drop $\bot$s from $A$ and $B$ to the x-axis and name them $AO$ and $BM$.
So, $tan(APO)$  = $tan(BPM)$
$\Rightarrow$ $\frac{1}{\Delta}$ = $\frac{2}{2-\Delta}$. 
Hence, $\Delta$=$\frac{2}{3}$
A: 
There are two solutions. If you only consider gradients (i.e. $\tan \alpha \text{ and } \tan \beta$), you will not find the other solution.
Intuitively, as P moves towards D $\beta$ decreases to 0, while $\alpha$ is increasing. Beyond D, both angles increase towards 90o, $\alpha$ faster than $\beta$, so there are no other solutions in that direction. However, there is another solution as P moves to the left beyond $\text{O}$.


*

*Consider the triangles AOP and DBP.

*Alternate angles between $\parallel$ lines are equal. So $\angle$PBD = ?, and $\angle$PAO = ?.

*Find values for PA and PB using Pythagoras' Theorem.

*Write expressions for $\sin \alpha$ and $\sin \beta$ in their respective triangles.

*For the angles to be the same, $\sin \alpha = \sin \beta$. Equate the two expressions from step 4, and solve for x.
