### Sept. 1, 2015, 5:26 p.m.

## A Determinant Deteremines the Line

### By Maurice Ticas

#### Tags:

A while back I helped a friend with his class assignment. In it contained the task to give the linear equation $$ax+by+c=0$$ for the line passing through the points \( (-2,-1), (0,4) \) expressed in slope-point form. I helped with providing the explanation:

The unique line passing through the two points \((-2,-1), (0,4)\) can be expressed as \(y=\frac{5}{2}x+4\). To see this, use the slope-intercept form of a line $$y=mx+b$$ where \(m\) is the slope and \(b\) a constant term. First calculate the slope from the two points; then choose one of the two points and plug it in for \(x,y\) to solve for the constant term \(b\). Here we chose the point \((0,4)\) to get a constant term value of \(4\).

The assignment also contained a word problem for which I also helped with the following explanation:

Let us say that a car three years later after purchase is worth $24,000 and that five years later it is worth $19,000. Assuming that depreciation is linear, we have that the two points \((3,24000),(5,19000)\)
uniquely determine the line \(5000x+2y-63000=0\). To get the slope-intercept form of the line, just solve for \(y\). The slope of the line will give the yearly rate of depreciation. The original price of the vehicle was $31,500.

The two above examples thus far might not feel new, but rather familiar and comfortable to grasp. What makes mathematics enjoyable is the ability to take what is familiar and see it from a different viewpoint. Let us now take the viewpoint of using a determinant.

The slope of a line can be readily read off from its linear form \(ax+by+c=0\) as \(-a/b\) assuming that \(b \neq 0\), i.e. the line isn't vertical. The problem now is just how to obtain this linear form of a line. To obtain a linear form of a line determined by two of its points \((x_1,y_1), (x_2,y_2)\)is computationally easy when one observes that the determinant
$$\left|
\begin{array}{ccc}
x & y & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{array}
\right|
$$
equal to \(0\) expresses our line up to a non-zero scalar multiple.

For those seeing determinants for the first time, it's best for our purposes here to just show how to calculate this determinant which expresses a line from two points. We'll not get caught up with formal definitions. All we need to do for our example of calculating this determinant is to add all signed factors of a given row or column multiplied with their respective co-factor.

To begin, we must determine the signs of the factors when calculating the determinant. So first decide on a row or column of factors. A decision on the row or column of factors determines the alternating signs used to calculate the determinant. The signs for a \(3 \times 3\) determinant are based on the table below:
$$\left|
\begin{array}{ccc}
+ & - & + \\
- & + & - \\
+ & - & +
\end{array}
\right|
$$
Usually the first row is chosen as the row of factors. Let us say we choose the first row. We then have \(+,-,+\) as our sequence of alternating signs for our factors \(x,y,1\). This then yields
$$+x \cdot c_1-y \cdot c_2+1 \cdot c_3=x \cdot c_1 -y \cdot c_2+c_3$$
as our determinant where \(c_1, c_2,\) and \(c_3\) are the co-factors of \(x,y,1\).

When calculating \(c_1,c_2\) and \(c_3\), we take its corresponding factor as a pivot element of the determinant taking out all entries from the column and row of the pivot element. The remaining entries determine a smaller \(2 \times 2\) determinant which becomes the co-factor . To illustrate,
$$c_1=
\left|
\begin{array}{cc}
y_1 & 1 \\
y_2 & 1
\end{array}
\right|
,c_2=
\left|
\begin{array}{cc}
x_1 & 1 \\
x_2 & 1
\end{array}
\right|
,c_3=
\left|
\begin{array}{cc}
x_1 & y_1 \\
x_2 & y_2
\end{array}
\right|
$$
where a \(2 \times 2\) determinant
\(
\left|
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right|\)
is defined as \(ad-bc\).

Putting all the pieces together, we then can finally see that
$$
0
=
\left|
\begin{array}{ccc}
x & y & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{array}
\right|=
x \cdot (y_1-y_2) -y \cdot (x_1-x_2) + (x_1 y_2-x_2 y_1)
$$

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