Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the line in $R^2$ that is given by the equation $d_1x_1 + d_2x_2 = c$ for numbers $d_1, d_2$ and $c$ in $R$ where $d_1$ and $d_2$ are not both zero. Find parametric equations of the line.

So I know a parametric equation for a line must have a form like:

$x_1 = T[d_1] + a_1$ and $x_2 = T[d_2] + a_2$ where $d_i$ is the direction vector and $a$ the translation vector. So, for the given equation I solve for $x_2$ like this:

$x_2 = \frac{c-d_1x_1}{d_2} = \frac{-d_1x_1}{d2} + \frac{c}{d_2}$ and start to see why are not both zero. But then, as one of them can be indeed zero, then it doesn't make sense to say that:

$x_1 = \frac{c-d_2x_2}{d_1}$

So I get stuck. I'm really trying hard to self-study linear algebra but this seems to much for me. Any help?

share|cite|improve this question
up vote 3 down vote accepted

This is just one way to do it: first get a points on the line by setting $x_1$ to zero:

$$d_1 \cdot 0 +d_2 x_2 = c$$ $$ x_2 = c/d_2$$

so $$(0,c/d_2)$$ is a point on the line. The point $(c/d_1,0)$ is also on the line, and can be obtained by setting $x_2 = 0$. The difference of these points (treated as vectors) is $(c/d_1,-c/d_2)$ which will be a vector parallel to the line.

The line can then be writen as $\vec{p_0} + t\vec{v}$ where $\vec{p_0}$ is a point on the line, $\vec{v}$ is a vector parallel to the line, and $t$ is the parameter. Therefore, one possible parametrization is:

$$(x,y) = (c/d_1,0) + t\cdot (c/d_1,-c/d_2).$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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