# Find two points on two lines in the plane where the line between the two points go through a third point and are equidistant from that point

I have the following situation (see pic below). I have two lines $B$, $C$, in the plane, the intersection point $a$, and a point $p$. I need to find the points $b$ and $c$ along $B$ and $C$ such that the line $A$ goes through $p$ and the two segments on each side of $p$ ($pb$ and $pc$) are of equal length, i.e., so that $l_1 = l_2$. I have tried setting up various equation systems with three unknowns (the lengths $|bc|$, $|ac|$, and $|ab|$), using the Cosine Law but I only get really complicated algebraic equations when trying to solve them.

Reflect $C$ in $p$, obtaining reflected line $C'$. Let $c$ be the intersection of $C'$ and $B$. I tried using analytic geometry, exploiting the fact that one can always situate $a$ at the origin and $p$ at $(0,1)$ in cartesian coordinates. Then the two lines $B$ and $C$ are simply of the form $y = \alpha_b x$ and $y = \alpha_c x$ and line $A$ is $y = \alpha_a x + 1$; we have solved the problem if we determine $\alpha_a$ in terms of $\alpha_b$ and $\alpha_c$ since we can alwyys rotate and rescale our original problem to fit this pattern.

Well, $$A \cap B = \left( \frac{1}{\alpha_b - \alpha_a},\frac{\alpha_b}{\alpha_b - \alpha_a} \right) \\ A \cap C = \left( \frac{1}{\alpha_d - \alpha_a},\frac{\alpha_c}{\alpha_d - \alpha_a} \right)$$ Now we write down the expressions for $|pa|^2$ and $|pb|^2$, and cross multiply by the product of the denominators, to get $$(\alpha_b-\alpha_a)^2 (1+\alpha_a^2) = (\alpha_c-\alpha_a)^2 (1+\alpha_a^2)$$ Then $$(\alpha_b-\alpha_a)^2 = (\alpha_c-\alpha_a)^2 \\ \alpha_a = \frac{\alpha_b+\alpha_c}{2}$$ Well, well, that is a pleasant surprise. THe geometric construction of line $A$ given point $p$ and lines $B$ and $C$ passing through point $a$ is now clear:

• Construct line $ad \perp pa$; $d$ can be any arbitrary point on this line other than $a$.
• Construct line $fd \perp ad$. Label the intersection of $df$ with line $B$ as $e$, and the intersection of $fd$ with line $C$ as $f$.
• Construct point $g$ as the midpoint of line segment $ef$.
• Draw line $ag$.
• Construct a line through $p$ which is outside line $ag$, parallel to line $ag$. Label that line $A$ because it is the desired line.

You can ensure equal distances by considering circles around point $P$ and then looking for the intersection points $B$ and $C$ with the lines $c$ and $b$.

(Note: I use upper case letters for points, lower case letters for lines).

Here are some of these: The interesting question is if we have a circle radius $R$ for which the vectors $u = PB$ and $v = PC$ are antiparallel $u = -v$.

The good news for this image is that there should be one, as the green and light blue cases show, there should be a feasible circle in between.

I do not know a good constructive way how to find that circle, in the above scenario one would try a circle and see if the lines along $PB$ and $PC$ are identical. Or one might look, if the line through $BC$ intersects with $P$.

Another analytical criteria would be that the area of the triangle $PBC$ vanishes.

• Sorry for not mentioning it in the question but I have considered this. It solves the problem but it requires an iterative solution and I was looking for a solution in constant time. Thanks anyway for your input.
– mags
May 12, 2015 at 6:52
• I prepared an analytic solution, but stopped after I saw Steven's solution which looked very simple. It seems to be constructable.
– mvw
May 12, 2015 at 6:58