# Draw a line l and two points A and B not lying on l. Make sure that the line bar(AB) is not perpendicular to l. Find the point C on l such that AC = BC?

Oct 13, 2016

See below.

#### Explanation:

Given a line

${L}_{1} \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1}$

and two points ${p}_{a}$ and ${p}_{b}$ not pertaining to ${L}_{1}$ and not perpendicular to ${L}_{1}$

for $p \in {L}_{1}$ we have

$\left\lVert p - {p}_{a} \right\rVert = \left\lVert p - {p}_{b} \right\rVert$ or

${\left\lVert {p}_{1} + {\lambda}_{1} {\vec{v}}_{1} - {p}_{a} \right\rVert}^{2} = {\left\lVert {p}_{1} + {\lambda}_{1} {\vec{v}}_{1} - {p}_{b} \right\rVert}^{2}$ or

${\left\lVert {p}_{1} - {p}_{a} \right\rVert}^{2} + 2 {\lambda}_{1} \left\langle{p}_{1} - {p}_{a} , {\vec{v}}_{1}\right\rangle + {\lambda}_{1}^{2} {\left\lVert {\vec{v}}_{1} \right\rVert}^{2} = {\left\lVert {p}_{1} - {p}_{b} \right\rVert}^{2} + 2 {\lambda}_{1} \left\langle{p}_{1} - {p}_{b} , {\vec{v}}_{1}\right\rangle + {\lambda}_{1}^{2} {\left\lVert {\vec{v}}_{1} \right\rVert}^{2}$ so

${\left\lVert {p}_{1} - {p}_{a} \right\rVert}^{2} + 2 {\lambda}_{1} \left\langle{p}_{1} - {p}_{a} , {\vec{v}}_{1}\right\rangle = {\left\lVert {p}_{1} - {p}_{b} \right\rVert}^{2} + 2 {\lambda}_{1} \left\langle{p}_{1} - {p}_{b} , {\vec{v}}_{1}\right\rangle$ so

${\lambda}_{1}^{\circ} = \frac{{\left\lVert {p}_{1} - {p}_{b} \right\rVert}^{2} - {\left\lVert {p}_{1} - {p}_{a} \right\rVert}^{2}}{2 \left\langle{p}_{b} - {p}_{a} , {\vec{v}}_{1}\right\rangle}$

The contact point in ${L}_{1}$ is

${p}_{c} = {p}_{1} + {\lambda}_{1}^{\circ} {\vec{v}}_{1}$

Attached an example with

${p}_{a} = \left(1 , 3\right)$
${p}_{b} = \left(4 , 3\right)$
${p}_{1} = \left(1 , 1\right)$ and
${\vec{v}}_{1} = \left(1 , 1\right)$

giving

${p}_{c} = \left(\frac{5}{2} , \frac{5}{2}\right)$