New x coordinate of a rotated line

Question

I need help finding the equation to find $x$

I work in GIS and I'm working on a script that uses the new x coordinate of a rotated line. I havent work with trigonometry in a long long time so I would much appreciate help with this equation.

I calculated the distance variable myself so it might be wrong $ \sqrt{(y_2 - y_1)^2 +(x_2-x_1)^2}$.

Answer 1

firstly the distance you have find between two known points,is wrong: $$\sqrt{(36000-35500)^2+(653000-653500)^2}\implies\sqrt{(500)^2+(-500)^2}\implies707.1\, approx.$$ then find its slope $m_2=\dfrac {653000-653500}{36000-35500}\implies m_2=-1$

suppose slope of 2nd line is $m_1$ then angle between two lines :

$$\tan \theta=|\dfrac{m_1-m_2}{1+m_1m_2}|$$ $$\tan 45^\circ=|\dfrac{m_1-m_2}{1+m_1m_2}|$$ $$1=|\dfrac{m_1+1}{1-m_1}|\implies m_1=0$$ it means this line is parallel to X-axis.So the coordinate $(x,y)$ will be $(x,653000)$. Now use distance formula (I assume that your given value $538.5m$ is right.)

$$\sqrt{(x-36000)^2}=538.5$$ $$x=36538.5m$$