Converting cartesian-form to vector-form line equations

Cartesian form

Cartesian form is the standard equation of a line, written as:

y = mx + c

(or one of its variants).

Vector form

A line in vector form has the equation:

\vec r = \vec a + \lambda \vec d

…where:

• \vec r is the position vector of any point on the line,
• \vec a is the position vector of a specific point on the line,
• \vec d is the direction vector of the line,
• \lambda is a scalar multiplier (changing it gives a new point on the line).

Direction vector

The direction vector has two components: the change in x and the change in y.

We can find these components from the gradient m of the line.

In short, the denominator of the gradient is the change in x, and the numerator is the change in y. (if there is no denominator, it is the same as a denominator of 1).

Find the direction vector from the gradient of \frac{3}{4}

• Change in x = 4
• Change in y = 3
• Answer: \vec d = \begin{pmatrix}4 \\ 3\end{pmatrix}

Find the direction vector from the gradient of -2

• Change in x = 1
• Change in y = -2
• Answer: \vec d = \begin{pmatrix}1 \\ -2\end{pmatrix}

Finding a point on the line

To find a specific point on the line, we can substitute any value of x into the cartesian equation to find the corresponding value of y.

However, the easiest point to find is the y-intercept, which is already stated by the cartesian line equation as the value of c.

For the equation y = mx + c, the y-intercept is at the point (0, c).

Converting from cartesian to vector form

Once we have the direction vector and a point on the line, we can substitute these into the vector form equation to get the final answer.

Convert the cartesian equation y = \frac{2}{3}x + 4 to vector form

• Gradient m = \frac{2}{3}
  • Change in x = 3
  • Change in y = 2
  • Direction vector \vec d = \begin{pmatrix}3 \\ 2\end{pmatrix}
• Y-intercept is at point (0, 4)
  • Position vector \vec a = \begin{pmatrix}0 \\ 4\end{pmatrix}
• Vector equation:
  • \vec r = \vec a + \lambda \vec d
  • = \begin{pmatrix}0 \\ 4\end{pmatrix} + \lambda \begin{pmatrix}3 \\ 2\end{pmatrix}
• Answer: \vec r = \begin{pmatrix}0 \\ 4\end{pmatrix} + \lambda \begin{pmatrix}3 \\ 2\end{pmatrix}

flashcards