1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $P\left(\dfrac{2}{3},\;0\right)$

$\endgroup$
0
$\begingroup$

enter image description here

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 $\text{O}$.

  1. Consider the triangles AOP and DBP.

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

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

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

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

$\endgroup$
0
$\begingroup$

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, $\Delta$=$\frac{2}{3}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.