Linear Equations

Lesson

Straight lines on the $xy$`x``y`-plane can actually be in any direction and pass through any two points. This means that straight lines can be:

decreasing |
horizontal |
||

increasing |
vertical |

Horizontal and vertical lines

We can quickly identify that a line is horizontal if it is parallel to the $x$`x`-axis. Similarly, a line is vertical if it's parallel to the $y$`y`-axis.

We can draw a set of points that makes a horizontal line in the same way we can draw any set of points as before. Consider the following table of values:

$x$x |
$-5$−5 | $-4$−4 | $-3$−3 | $-2$−2 |
---|---|---|---|---|

$y$y |
$2$2 | $2$2 | $2$2 | $2$2 |

Plotting each column as points on the $xy$`x``y`-plane gives us the following:

Plotting points from the table of values |

Clearly the set of points form a line that is parallel to the $x$`x`-axis. But just to confirm, we can draw a line through these points to show this.

Horizontal line passing through all four points |

It's also clear from the graph that the distance between any two neighbouring points is exactly $1$1 unit apart. We obtain the distance by referencing the $x$`x`-axis, and subtracting the $x$`x`-values of the two points. This means we don't have to plot the points to determine the distance between a pair of points. Consider the table of values earlier.

$x$x |
$-5$−5 | $-4$−4 | $-3$−3 | $-2$−2 |
---|---|---|---|---|

$y$y |
$2$2 | $2$2 | $2$2 | $2$2 |

The distance between the points $\left(-4,2\right)$(−4,2) and $\left(-3,2\right)$(−3,2) is:

$-3-\left(-4\right)=1$−3−(−4)=1 unit.

Careful!

You may find yourself with a negative distance depending on what order you subtract the two $x$`x`-values. Make sure to subtract in the correct order or change your final answer to a positive number.

Just as we did before, we can plot a set of points obtained from a table of values and show that the points fall on a vertical line. Consider the following table of values:

$x$x |
$2$2 | $2$2 | $2$2 | $2$2 |
---|---|---|---|---|

$y$y |
$-3$−3 | $-1$−1 | $1$1 | $3$3 |

Plotting the points in the table of values gives us the following graph. We can also pass a line through the points to show that the points lie on a vertical line.

Plotting points from the table of values |

By referencing the $y$`y`-axis we can see that the distance between a pair of neighbouring points is $2$2 units. Alternatively we can refer to the ordered pairs $\left(2,-1\right)$(2,−1) and $\left(-1,-3\right)$(−1,−3) and subtract the two $y$`y`-values:

$-1-\left(-3\right)=2$−1−(−3)=2 units

We still need to be careful about the order we subtract the two $y$`y`-values to ensure we obtain a positive distance (or you could change your final answer so that it's positive).

Let's have a look at some examples of questions that might arise when dealing with horizontal or vertical lines.

Plot the points in the table of values.

$x$x |
$2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|

$y$y |
$5$5 | $5$5 | $5$5 | $5$5 |

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Consider the points in the plane below.

Which of the following statements is true?

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The set of points lie on a decreasing line.

AThe set of points lie on an increasing line.

BThe set of points lie on a vertical line.

CThe set of points lie on a horizontal line.

D

What is the shortest distance between any two of the following points?

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