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I have a two points, $A(x_1,y_1)$ and $B(x_2,y_2)$. They are united with a line.

Now I have a 3rd and 4th point $C(x_3,y_3)$, $D(x_4,y_4)$. The line formed by CD is going to intercept AB in point $E(x_5,y_5)$.

I know everything except $y_3, y_4$ and $y_5$, also $x_3, x_4, x_5$ are equal.

How do I find that $y_3,y_4,y_5$? Is there a formula?

Thanks.

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Point $E=(x_5,y_5)$ is the intersection of the line $AB$ with the $CD$ (which we know to be perpendicular to the X-axis because $x_3=x_4$). So a formula for the line $AB$ is $$(x_1-x_2)y= (y_1-y_2)x + y_1(x_1-x_2) - x_1(y_1-y_2)$$

and so $$y_5= \frac{(y_1-y_2)x_5 + y_1(x_1-x_2) - x_1(y_1-y_2)}{x_1-x_2}$$

You can't find $y_3$ and $y_4$ because the only thing you know about points $C$ and $D$ is that they lie in the same vertical line and the construction does not change if you move them along that same line (try drawing it).

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  • $\begingroup$ This actually came from a drawing. I'll test it out later, thanks a lot for now. I was testing to see if I can make a line AB jump on a randomly generated line CD, sort of like in a platforming game, but on canvas and without using foreign libraries - I hit that bump. $\endgroup$ – Andrei Cristian Prodan Jan 15 '13 at 12:54

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