### Introduction to Linear Algebra

#### Linear Equations

##### Equations of a Line

There are various equations of a line, all derivable from each other. In the previous lesson o we covered the slope-intercept form of the equation of a line:

y = mx + b

where m is the slope, and b is the y-intercept.

Some countries use different symbols to denote slope and y-intercept. For example, in Sweden, slope is denoted k and y-intercept is denoted m. In that case, the slope-intercept equation of a line is:

y = kx + m

Following are examples of lines in slope-intercept form: Figure 1: Slope-intercept line equations. [JimBelk]

The line y = −x + 5 has slope −1, and y = (½)x + 2 has slope 1/2.

Notice that in the positive-x direction (+ →), slope is positive uphill (+ ↑) and negative downhill (− ↓). That is because slope is “rise over run”, and rise is positive uphill, negative downhill.

The y-intercepts of those lines are y = 5 and y = 2. If the terms +5 and +2 respectively were removed from those equations, both lines would pass through the origin (would have y-intercept y = 0). The y-intercepts are essentially vertical offsets of the lines, away from the origin.

Those two lines would intersect at the origin if the y-intercepts (vertical offsets) of the lines were both zero: Figure 2: Lines that both have y-intercept equal to zero intersect at the origin.

For lines that have the same y-intercept, the intersection of those lines would be on the y-axis (at x = 0): Figure 3: Lines that have the same y-intercept intersect on the y-axis (at that y-intercept).

The intersection of those lines is offset vertically from the origin, as both lines are offset vertically by the same amount. In that case, the intersection of the lines is not offset horizontally.

The intersection of the lines is offset vertically but not horizontally, because the lines have the same y-intercept. This is regardless of the slopes being different.

If the y-intercepts of the lines are different (not the same as just described), then the intersection of the lines will be offset from the y-axis: Figure 4: The intersection of two lines will not be on the y-axis if the y-intercepts of the lines are different.

##### General Form

Besides the slope-intercept form of the equation of a line, a more general equation of a line is:

ax + by + c  =  0

where a, b and c are coefficients. Using those coefficients, the slope of the line is −(a/b), and the y-intercept is −(c/b).

This is called the “general form” (or “standard form”) of the equation of a line. The coefficients are often denoted as capital letters:

Ax + By + C  =  0

Using those coefficients, the slope of the line is −(A/B), and the y-intercept is −(C/B).

Another general (standard) form of the equation of a line is:

Ax + By  =  C

In that case, the slope of the line is −(A/B), and the y-intercept is C/B.

###### Exercise

Problem: Convert the equations, of the lines we have been discussing, into general form Ax + By + C = 0.

Solution:

𝑦
𝑥 + 𝑦
𝑥 + 𝑦 − 5
−𝑥 + 5
=  5
=  0

where A = 1, B = 1, and C = −5.

And:

𝑦
−(½)𝑥 + 𝑦
−(½)𝑥 + 𝑦 − 2
(½)𝑥 + 2
=  2
=  0

where A = −1/2, B = 1, and C = −2.

##### Simultaneous Equations

Converting the line equations we have been discussing, to the second general form of the equation of a line, yields this system of equations:

𝑥 + 𝑦
−(½)𝑥 + 𝑦
=  5
=  2

These equations together (a “system”), can be solved simultaneously to find the point of intersection of the two lines. We refer to such a system of equations as “simultaneous equations”.

One way to solve these equations, to find the intersection point, is to use algebraic substitution: we can substitute one equation into the other equation to find one of the point coordinates, then substitute that coordinate into either equation to find the other coordinate.

For example, converting the first equation to x = 5−y, then substituting that for x in the second equation:

𝑥 + 𝑦
𝑥

−(½)𝑥 + 𝑦
−(½)(5−𝑦) + 𝑦
−5/2 + 𝑦/2 + 𝑦
𝑦/2 + 𝑦
𝑦/2 + (2/2)𝑦
(3/2)𝑦
3𝑦
𝑦
=  5
=  5 − 𝑦

=  2
=  2
=  2
=  2 + 5/2
=  4/2 + 5/2
=  9/2
=  9
=  3

giving the y-coordinate (3) of the intersection point. Plugging that into x = 5−y produces the x-coordinate 2. Hence the intersection point is (2,3).

While algebraic substitution is possible for simple lines like this, you will eventually work with lines that cannot be solved only with substitution. We now turn to another method of solving simultaneaous equations.

Solving simultaneous equations to find the intersection of the equations can also be accomplished by replacing one or more of the equations with simpler equations that have the same intersection, and then solving the simpler equations instead.

Consider, again, the graph of the two lines we have been discussing: Figure 5: Two intersecting lines.

If you add the two line equations together, the resulting equation produces a line that also intersects those lines at the same point of intersection:

[       x +  y = 5 ]
+ [ -(1/2)x +  y = 2 ]
________________________

= [  (1/2)x + 2y = 7 ] Figure 6: The red line is produced by adding the equations of the black lines together.

Note also that scaling a line equation produces the same line. For example, x+y=5 produces the same line as 2x+2y=10, 3x+3y=15, −3x−3y=−15, etc. Therefore, subtracting line equations also produces a line that intersects the lines, since the subtraction is addition of a scaled equation (that was scaled by a negative number).

We subtract the equations of the lines we have been discussing:

[       x +  y = 5 ]
- [ -(1/2)x +  y = 2 ]
________________________

= [  (3/2)x + 0y = 3 ]
= [       x      = 2 ] Figure 7: Red line produced by subtracting the equations of the black lines.

This subtraction of equations was useful in “eliminating” one of the variables (y), directly producing the x-coordinate of the intersection point (x = 2), which can be plugged into any of the other equations of this system to produce the y-coordinate of the intersection point.

Question: Why are all lines, passing through an intersection point, multiples and sums of each other?

Answer: Because the intersection point is like a “new origin”, with those lines forming a “space” there. That space is referred to as a “vector space”.

Graphs and Lines
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