Equation of a line:
$$Ax+By+C=0$$
Point on this line:
$$D_1(x_1, y_1)$$
How to find coordinates of points $D_2(x_2, y_2)$ and $D_3(x_3, y_3)$ lying at distance $r$ from point $D_1$ on this line?
3
-
$\begingroup$ By "distance r" you mean the distance normal to the line? $\endgroup$– AndreasJan 2, 2018 at 13:18
-
$\begingroup$ Question seems a bit vague $\endgroup$– QuIcKmAtHsJan 2, 2018 at 13:19
-
$\begingroup$ Draw a line then draw point $D_1$ on this line. Count 3 ($r=3$) units to the left and right from $D_1$ and put the points $D_2$ and $D_3$ on this line. Their coordinates I want to find $\endgroup$– ValeriyJan 2, 2018 at 13:23
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
If the distance $r$ from $D_1$ is meant normal to the line, you have points (x,y) with
(distance)
$ (x-x_1)^2 + (y-y_1)^2 = r^2 $
For normality: the line $Ax+By+C=0$ has slope $-A/B$. The vector $(y-y_1, x-x_1)$ is normal to the line, so it must have slope $B/A$:
$ \frac{x-x_1}{y-y_1} = A/B $
From here the two solutions for $D_2$ and $D_3$ follow.
$\begingroup$
$\endgroup$
$$y_i=-\dfrac{Ax_i+C}B$$
$$r=\sqrt{(x_1-x_i)^2+\left(\dfrac{A(x_i-x_1)}B\right)^2}=\dfrac{|x_i-x_1|\sqrt{A^2+B^2}}{|B|}$$ where $i=2,3$
$$\implies x_i-x_1=\pm\dfrac{Br}{\sqrt{A^2+B^2}}$$
Clearly, $x_1,A,B,r$ are known
