# Implicit Equations for Lines

So, I’ve covered parametric lines already. Another form in which we can write equations for lines using linear algebra is implicit form.

The parameter in the parametric form of a line was a scalar $$\mathtt{k}$$. We built the parametric form using a position vector to get us to a starting point on the line. Then we added this to the product of the slope vector and the parameter $$\mathtt{k}$$ to get all the other points on the line. The implicitness of the implicit form comes from the fact that we build the equation using the slope vector and a vector perpendicular to the slope vector.

I mentioned back here that perpendicular vectors always have a dot product of 0. So, thinking of $$\mathtt{x-p}$$ as the (slope) vector of our line, then $$\mathtt{a \cdot (x-p) = 0}$$. With the actual values shown here, we have $\begin{bmatrix}\mathtt{1}\\\mathtt{3}\end{bmatrix} \cdot \begin{bmatrix}\mathtt{\,\,\,\,3}\\\mathtt{-1}\end{bmatrix}\mathtt{ = (1)(3) + (3)(-1) = 0}$ If we know one point $$\mathtt{x}$$ on the line, then the dot product equation is true for any $$\mathtt{p}$$ and identifies a unique line. Let’s represent all the parts here as vectors, and more generally. $\begin{bmatrix}\mathtt{a_1}\\\mathtt{a_2}\end{bmatrix} \cdot (\begin{bmatrix}\mathtt{x_1}\\\mathtt{x_2}\end{bmatrix} \mathtt{- }\begin{bmatrix}\mathtt{p_1}\\\mathtt{p_2}\end{bmatrix}) \mathtt{\,\,= 0 \rightarrow a_1x_1 + a_2x_2 + (-a_1p_1 – a_2p_2) = 0}$

What’s cool about this equation is that we are all familiar with its form, $$\mathtt{ax_1 + bx_2 + c = 0}$$, so long as we let $$\mathtt{a = a_1, b = a_2,}$$ and $$\mathtt{c = -a_1p_1 – a_2p_2}$$. This is what is called the general form or standard form of a linear equation. Even more interesting is that the coefficients in this form help to describe a vector perpendicular to the line.

Knowing the above and $$\mathtt{y:=x_2}$$, we can write the equation for the line at the right as $\mathtt{4x+3y-(4)(-5)-(3)(6)=0}$ And then, working out that c-value, we get $$\mathtt{4x + 3y + 2 = 0}$$. The vector a = (4, 3), which we can rewrite as the ratio –4 : 3, describes the slope of the line.

Now we can easily slide back and forth between linear algebra and plain old current high school algebra with certain linear equations.

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