0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ By "distance r" you mean the distance normal to the line? $\endgroup$ – Andreas Jan 2 '18 at 13:18
  • $\begingroup$ Question seems a bit vague $\endgroup$ – QuIcKmAtHs Jan 2 '18 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 $\endgroup$ – user1943863 Jan 2 '18 at 13:23
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks. Can you add some explain how it was given? I well understand only "distance" part. $\endgroup$ – user1943863 Jan 2 '18 at 13:37
  • $\begingroup$ For normality: I put the explanation in the text. $\endgroup$ – Andreas Jan 2 '18 at 13:46
1
$\begingroup$

$$y_i=-\dfrac{Ax_i+C}B$$

$$r=\sqrt{(x_1-x_i)^2+\left(\dfrac{A(x_i-x_1)}B\right)^2}=\dfrac{|x_i-x_1|\sqrt{A^2+B^2}}{|B|}$$ 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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.