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i have two lines, one is slanted line and another is a straight line. Both the lines share same X coordinate separated by distance d. The straight line is rotated to an angle θ (which also needed to be calculated from slanted line) ,making it parallel to the slanted line. Now how to calculate the new end points X2,Y2 of the rotated line?

  • I used the slope equation to find the m1 of the slanted line and then converted it into angle using m=tanθ formula.
  • Now how to calculate the new X2,Y2 after rotating the line?

enter image description here

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  • $\begingroup$ You may want to use $(x,y)=(r\cos\theta,r\sin\theta)$. $\endgroup$ Commented Apr 11, 2020 at 8:34

2 Answers 2

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No need to compute any angles or trigonometric functions of them directly. Since you want to end up parallel to the second line segment after rotation, you already have a vector that’s pointing in the right direction, namely $P_4-P_3$. You just need to adjust its length and add it to $P_0$: $$P_2=P_0+{\lvert P_1-P_0\rvert\over\lvert P_4-P_3\rvert}(P_4-P_3).$$

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  • $\begingroup$ Thanks for the reply. So the P0 u mentioned above is it the points X0,Y0 + distance ? but what distance? the distance between the lines or the distance between the P0 and P2? $\endgroup$
    – Balaram26
    Commented Apr 11, 2020 at 9:22
  • $\begingroup$ @Balaram26 The points $P_i$ are numbered just as you have them, e.g., $P_0=(x_0,y_0)$. $\endgroup$
    – amd
    Commented Apr 11, 2020 at 18:16
  • $\begingroup$ will the equation be the same even when the line order changes as line P3,P4 comes first and line P0,P1,P2 comes second ? $\endgroup$
    – Balaram26
    Commented Apr 12, 2020 at 12:20
  • $\begingroup$ @Balaram26 I have no idea what that means. How does a line segment “come first?” The formula works no matter how these points are laid out. $\endgroup$
    – amd
    Commented Apr 13, 2020 at 0:00
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Let $AB$ the segment from $(x_0,y_0)$ to $(x_1,y_1)$. Note that when you rotate the segment $AB$ this point that we call $P$ rotates, describing an arc of angle $\theta$.

So, let the origin $O(x_0,y_0)$, and draw the perpendicular line to the segment $AB$ from $S(x_2,y_2)$. Let $K$ the intersection, the triangle $SOK$ is a right triangle. So: $$x_2=|x_1|\cos(\theta)$$ and: $$y_2=|x_1|\sin(\theta)$$ Note that we can write $|x_1|$ bevause we assume that the segment $AB$ is part of $x-$axis.

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  • $\begingroup$ Is it typo or u mean the segment moved from (x0,y0) to (x1,y1) ? because i am moving the (x1,y1) to (x2,y2) $\endgroup$
    – Balaram26
    Commented Apr 11, 2020 at 9:23
  • $\begingroup$ @Balaram26: please see my edits. $\endgroup$
    – Matteo
    Commented Apr 11, 2020 at 9:36
  • $\begingroup$ @Balaram26: if this answer has been useful, maybe a $+1$ or a favourite answer? $\endgroup$
    – Matteo
    Commented Apr 11, 2020 at 10:09

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