Equation of a line:


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?

  • $\begingroup$ By "distance r" you mean the distance normal to the line? $\endgroup$
    – Andreas
    Jan 2, 2018 at 13:18
  • $\begingroup$ Question seems a bit vague $\endgroup$
    – QuIcKmAtHs
    Jan 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 $\endgroup$
    – Valeriy
    Jan 2, 2018 at 13:23

2 Answers 2


If the distance $r$ from $D_1$ is meant normal to the line, you have points (x,y) with


$ (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.

  • $\begingroup$ Thanks. Can you add some explain how it was given? I well understand only "distance" part. $\endgroup$
    – Valeriy
    Jan 2, 2018 at 13:37
  • $\begingroup$ For normality: I put the explanation in the text. $\endgroup$
    – Andreas
    Jan 2, 2018 at 13:46


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


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