Problem on finding a line between two points
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Find the line containing the points $(0, 1)$ and $(3, 0)$ by computing a normal vector to the line. Write it in the form $\bfn \cdot \bfx = b$.
Solution
Recall that
Lines in 2d can be specified by a point and a normal vector
$\bfn \cdot \bfx = b$ is the equation of a line in 2d with normal vector $\bfn$.
If the line contains $\bfx_0$ then $b = \bfn \cdot \bfx_0$.
We know the line contains $(0,1)$, so we choose $\bfx_0 = (0,1)$.
We now look for a normal vector $\bfn$.
Because $\bfn$ is perpendicular to the line, it must be perpendicular to the vector from $(0,1)$ to $(3,1)$.
Subtraction gives the vector between two points
The vector from $(0,1)$ to $(3,0)$ is $(3, -1)$.
Thus, we are looking for a vector perpendicular to $(3,-1)$.
Recall that
Finding a perpendicular vector
A vector perpendicular to $ \langle a,b \rangle$ is $\langle -b, a \rangle$.
So, we take $\bfn = (1, 3)$.
We now compute that $b = \bfn \cdot \bfx_0 = (1,3) \cdot (0,1) = 3$.
The line containing the points $(0,1)$ and $(3,0)$ is $$(1,3) \cdot (x,y) = 3.$$
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