The line between algebra and geometry is gray and blurry—the two mesh a great deal. “Linear algebra” and “planar geometry” are truly close cousins. The first concept in the “Geometry Study Hall” is as much algebra as it is geometry. But, that’s okay! One helps the other. Let’s first look at the linear equation.

## Graphing linear equations

The formula for the “linear equation” or the “equation for a line” is usually shown in what’s called the “slope-intercept form.” It appears as:

$y=mx+b$

m = the slope

b = the y-intercept (the point where the line crosses the y axis)

Note that the equations given on a test usually aren’t in this form initially. You often have to manipulate the equation to get it in this form. It may be shown as something like: $5y–2x=10$. This can easily be changed to the slope-intercept form of the linear equation by isolating the y on one side of the equals sign. If you do so, you’ll wind up with: $y= {2 \over 5} x+2$. This is now in the $y=mx+b$ format.

But, how do you graph a line in this form? Take the linear equation below as an example:

$y=-2x+1$

Graphing this equation is easy. Start with the y-intercept of 1, mark a point there. Then use the “rise over run” principle of slope. The slope is –2, which is to say it’s ${-2\over1}$ So, you’ll “rise” –2 points (you actually “rise” downward here because of the negative sign) and then “run” to the right (because it’s positive) 1 unit. Mark a second point there. Connect your points to draw your line. See the graphic below for illustration purposes.

## Finding the linear equation through two points

If you’re given two points, you can determine the linear equation. Consider the two points: (2,4) and (3,7). You could easily draw these points on graphing paper, draw the line, and visually look and figure out the formula. Or, you can solve it algebraically.

The first thing to do is to find the slope. Slope is usually marked with an $m$. The formula for finding the slope is:

$m={y_2-y_1\over x_2-x_1}$

Plugging in the numbers from the points gives us:

$m={7-4 \over 3-2}$ which becomes $m=3 \over 1$ So, the slope $m = 3$.

Using the slope-intercept form of the linear equation and knowing that 2 is the slope, we now have: Using the slope-intercept form of the linear equation and knowing that 2 is the slope, we now have:

$y=3x+b$

The next step is to find the y-intercept (b). To do that, simply plug in the $x$ and $y$ coordinates from one of the points and solving for b. We’ll use (2,4). So, now the equation reads:

$4 = 3(2) + b$

Which is simplified to:

$4 = 6 + b$

Which is simplified to:

$-2 = b$

The y intercept is –2. So, our final equation reads:

$y = 3x - 2$

## Finding the intersection of two lines

Given two lines (in some form of a linear equation) you should be able to find the point at which those lines intersect. First things first, however—a couple of questions might come up here…

- What if the lines don’t intersect?
- How about perpendicular lines? Anything special here?

### Parallel lines

All lines in a plane must eventually intersect, unless they’re parallel. You can easily recognize parallel lines because they share the **same slope**.

### Perpendicular lines

Lines that intersect perpendicularly (at right angles) are also easily recognized. They have **negative reciprocal** slopes. That is to say if one slope is 2, the other slope is $-{1 \over 2}$. Or another example, if one slope is $-{5 \over 2}$, then its perpendicular counterpart has a slope of ${2 \over 5}$. The slopes are the positive/negative opposite, and they’re inverted (flipped).

Finding the intersection of two lines can be done either graphically or algebraically. We’ll do both.

## Graphical solution

To graphically find the intersection of two lines, you simply draw them. Consider the following two lines: line K $x+2y=5$ and line L $2x–y=0$ We’ll graph these two lines. However, they’re not written in slope-intercept form. So, step one is to put them in that form so they can be easily graphed. To change the lines above to slope-intercept form, simply solve each equation for y.

Line K: $x+2y=5$ becomes ${2y \over 2}={-x+5 \over 2}$ which in turn becomes $y=-{1 \over 2}x+2.5$ This is now in slope-intercept form.

Line L: $2x-y=0$ becomes $-y=2x$ If you multiply both sides by -1, the final equation becomes $y=-2x$. This is now in slope-intercept form of $y=mx+b$ (Note, the $+b$ part of the formula is really just "plus 0" so it’s not written).

The graphic shows the point of intersection at (1,2). That seems simple enough, but there’s a weakness here. How can you be sure that it’s (1,2) and not, say (1, 2.06)? Just by *looking*, you really *can’t know for certain*. So, you must find the intersection point using algebra.

## Algebraic solution

To find the intersection algebraically, go back to the initial equations:

Line K: $x+2y=5$

Line L: $2x–y=0$

These equations have two variables in each one. We need to get that down to one variable. To do that, refer to the graphic below: