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How to find the point of intersection of two lines, given four points, two of which are on each line, in complex numbers?

Thank you!

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closed as off-topic by Travis, Crostul, MathOverview, user98602, Shaun Jul 7 '15 at 17:39

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  • $\begingroup$ Write the equation of the two lines and then solve them to get the desired point. $\endgroup$ – MathGod Jul 7 '15 at 12:49
  • $\begingroup$ How do you write the equation of a line through two points in complex numbers? $\endgroup$ – Bob Joe Jul 7 '15 at 12:53
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The condition that $z$ is on the line which passes through two points $z_1,z_2$ is $$\frac{z-z_1}{z_2-z_1}=\overline{\left(\frac{z-z_1}{z_2-z_1}\right)},$$ i.e. $$(z-z_1)(\overline{z_2}-\overline{z_1})=(z_2-z_1)(\overline{z}-\overline{z_1}),$$ i.e. $$(\overline{z_2}-\overline{z_1})z-(z_2-z_1)\overline{z}=(\overline{z_2}-\overline{z_1})z_1-(z_2-z_1)\overline{z_1}\tag1$$ Similarly, the condition that $z$ is on the line which passes through two points $z_3,z_4$ is $$(\overline{z_4}-\overline{z_3})z-(z_4-z_3)\overline{z}=(\overline{z_4}-\overline{z_3})z_3-(z_4-z_3)\overline{z_3}\tag2$$

Multiplying the both sides of $(1)$ by $(z_4-z_3)$ gives $$(z_4-z_3)(\overline{z_2}-\overline{z_1})z-(z_4-z_3)(z_2-z_1)\overline{z}=((\overline{z_2}-\overline{z_1})z_1-(z_2-z_1)\overline{z_1})(z_4-z_3)\tag3$$ Multiplying the both sides of $(2)$ by $(z_2-z_1)$ gives $$(z_2-z_1)(\overline{z_4}-\overline{z_3})z-(z_4-z_3)(z_2-z_1)\overline{z}=((\overline{z_4}-\overline{z_3})z_3-(z_4-z_3)\overline{z_3})(z_2-z_1)\tag4$$

Now $(3)-(4)$ gives $$(z_4-z_3)(\overline{z_2}-\overline{z_1})z-(z_2-z_1)(\overline{z_4}-\overline{z_3})z$$$$=((\overline{z_2}-\overline{z_1})z_1-(z_2-z_1)\overline{z_1})(z_4-z_3)-((\overline{z_4}-\overline{z_3})z_3-(z_4-z_3)\overline{z_3})(z_2-z_1),$$ i.e. $$z=\frac{((\overline{z_2}-\overline{z_1})z_1-(z_2-z_1)\overline{z_1})(z_4-z_3)-((\overline{z_4}-\overline{z_3})z_3-(z_4-z_3)\overline{z_3})(z_2-z_1)}{(z_4-z_3)(\overline{z_2}-\overline{z_1})-(z_2-z_1)(\overline{z_4}-\overline{z_3})}$$

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