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I'm trying to find the last point in a triangle like this below enter image description here

I know what $X1,Y1$ and $X2,Y2$ is. Now i need to Find $X3,Y3$ which should lay at a angle of $45°$ from $X2,Y2$

I can get the length between $X1Y1$ And $X2,Y2$ If needed.

I know this should be a simple matter, but i just can't find the right way to do it. And by searching online, i only found some high level math, i could not follow. So please bare with me, an explain how this can be done, for stupid me.

Edu
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DaCh
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  • The length between $(X_1,Y_1)$ and $(X_2,Y_2)$ is equal to the length between $(X_1,Y_1)$ and $(X_3,Y_3)$. And you mean at an angle of $45^o$ from $(X_1,Y_1)$ not $(X_2,Y_2)$. – Jimmy R. Oct 14 '16 at 08:42
  • As you can see, i'm not good ad math. but with this, i still don't know how to find the last point. – DaCh Oct 14 '16 at 08:43

2 Answers2

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Using complex numbers

\begin{align*} \frac{z_3-z_1}{z_2-z_1} &= \operatorname{cis} \frac{\pi}{4} \\ &= \frac{1+i}{\sqrt{2}} \\ z_3 &= z_1+\frac{(1+i)(z_2-z_1)}{\sqrt{2}} \\ x_3+y_3 i &= x_1+y_1 i+\frac{1+i}{\sqrt{2}} [(x_2-x_1)+(y_2-y_1)i] \\ &= \left( x_1+\frac{x_2-x_1+y_1-y_2}{\sqrt{2}} \right)+ \left( y_1+\frac{x_2-x_1+y_2-y_1}{\sqrt{2}} \right)i \\ \begin{pmatrix} x_3 \\ y_3 \end{pmatrix} &= \begin{pmatrix} x_1+\frac{x_2-x_1+y_1-y_2}{\sqrt{2}} \\ y_1+\frac{x_2-x_1+y_2-y_1}{\sqrt{2}} \end{pmatrix} \end{align*}

Only with elementary backgrounds

Let $A=(x_1,y_1)$, $B=(x_2,y_2)$, then

$$C=(x_2+y_1-y_2 ,\, y_2+x_2-x_1)$$

Note that $$AD : DC = 1: \sqrt{2}-1$$

By section formula

\begin{align*} D &= \frac{C+(\sqrt{2}-1)A}{1+(\sqrt{2}-1)} \\ &= \frac{(x_2+y_1-y_2 ,\, y_2+x_2-x_1)+(\sqrt{2}-1)(x_1 ,\, y_1)} {\sqrt{2}} \\ (x_3, y_3) &= \left( x_1+\frac{x_2-x_1+y_1-y_2}{\sqrt{2}} ,\, y_1+\frac{x_2-x_1+y_2-y_1}{\sqrt{2}} \right) \end{align*}

enter image description here

Ng Chung Tak
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  • I'm sure this is awesome, But as I said, I'm very stupid at math, So I kinda don't get what is going on here. – DaCh Oct 14 '16 at 10:31
  • Rotate from $(x_2,y_2)$ by $45^{\circ}$ anti-clockwise about $(x_1,y_1)$, you reach $(x_3,y_3)$. – Ng Chung Tak Oct 14 '16 at 10:36
  • Sry, i didnt explain myself properly, I can't read the formular, is like a other language. I do understand that if i can rotate it 45 degree, it would be at that point. but can't understand the formular for doing the math – DaCh Oct 14 '16 at 10:45
  • Have you learnt complex numbers (that involving $\sqrt{-1}$)? If not, how about matrix algebra? – Ng Chung Tak Oct 14 '16 at 10:45
  • To some point. But what is i, And what is cis? that i do not know. I think what i'm missing here is a step by step explanation of what this formular do – DaCh Oct 14 '16 at 11:12
  • $i=\sqrt{-1}$ and $\operatorname{cis} \theta=\cos \theta+i\sin \theta$ – Ng Chung Tak Oct 14 '16 at 11:14
  • This is what i need, Now i understand, Thanks a lot. – DaCh Oct 14 '16 at 11:55
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1968054

$x_3=x_1+\frac{(x_2+y_1-x_1-y_2)\sqrt2}2$
$y_3=y_1+\frac{(x_2+y_2-x_1-y_1)\sqrt2}2$

I derived these from the following formulæ:
$\left\{\begin{matrix} X=\frac{(x_2-x_1)(x-x_1)+(y_2-y_1)(y-y_1)}{(x_2-x_1)^2+(y_2-y_1)^2}\\ Y=\frac{-(y_2-y_1)(x-x_1)+(x_2-x_1)(y-y_1)}{(x_2-x_1)^2+(y_2-y_1)^2} \end{matrix}\right.$
These map $(x_1\mid y_1)\text{ onto }(0\mid 0)\text{, }$
$(x_2\mid y_2)\text{ onto }(1\mid 0)\text{ and other points accordingly.}$

$\left\{\begin{matrix} x=(x_2-x_1)X-(y_2-y_1)Y+x_1\\ y=(y_2-y_1)X+(x_2-x_1)Y+y_1 \end{matrix}\right.$
These reverse the mapping.

Senex Ægypti Parvi
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  • Will you share of what significance is the highlighted 7-digit number? – amWhy Oct 15 '16 at 22:25
  • @amWhy \ Gladly! That's the number that MathSE assigned to the question. I include it with my answer just to keep my mind straight, particularly when I intend to return to the question repeatedly until I am satisfied. – Senex Ægypti Parvi Oct 15 '16 at 22:32
  • Ahhh, that makes sense to do! Thanks for sharing! – amWhy Oct 15 '16 at 22:36