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Given the following graph:

enter image description here

Where SP is the start point and EP is the end point of a line - how would I rotate p1 and p2 on axis SP in such a way that they would represent an isosceles triangle e.g.

enter image description here

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What is special about the second picture? For example, do you want your triangle to be isosceles? – dtldarek Feb 4 '13 at 9:03
@dtldarek Judging from the title of the question, I'd say 'yes' – mrf Feb 4 '13 at 9:10
@dtldarek yes, I have updated the question. – James Feb 4 '13 at 9:26

Calculate the distance between say $p_1$ and $SP$, $d = |SP - p_1|$. Suppose $v := EP - SP = (v_1, v_2)$. Then $n = (-v_2, v_1)$ is the direction you need to go in. Then your points are $SP \pm \frac{d n}{|n|}$

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Could you elaborate a little on v := EP - SP = (v1, v2)? i.e. where does v1/v2 come from (bit rusty with maths!). – James Feb 4 '13 at 10:16
Let $EP = (x_1, y_1)$ and $SP = (x_2, y_2)$. Then what I meant is that $v = (x_2 - x_1, y_2 - y_1)$. So $n$ would then be $(y_1 - y_2, x_2 - x_1)$. – muzzlator Feb 4 '13 at 10:18
Thanks and the last formula could you explain the last equation i.e. dn/|n|? I understand that p1/p2 will be the SP +- whatever the value of that equation is. – James Feb 4 '13 at 10:29
I suppose you need to know what $|v|$ means, $|v|$ is the length of a vector $v$ and you can calculate it in 2-dimensions by doing $|(v_1, v_2)| = \sqrt{v_1^2 + v_2^2}$. So, after you find $d = |SP - p_1|$, you then take the vector we just calculated $n$, first scale it by $d$, then scale it by $1 / |n|$ – muzzlator Feb 4 '13 at 10:39
So in laymans terms - P1 := SP - (d x n / √ n)? – James Feb 4 '13 at 10:49

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