Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

To graph any liner relationship you only need two points that are on the line. You can use any two points from a table of values, or substitute in any two values of $x$`x` into the equation and solve for corresponding $y$`y`-value to create your own two points. Often, using the intercepts is one of the easiest ways to sketch the line.

x | 1 | 2 | 3 | 4 |

y | 3 | 5 | 7 | 9 |

To sketch from a table of values, we need just any two points from the table. From this table we have 4 coordinates, $\left(1,3\right)$(1,3), $\left(2,5\right)$(2,5), $\left(3,7\right)$(3,7), $\left(4,9\right)$(4,9).

Drag the $2$2 of the points on this interactive to the correct positions and graph this linear relationship.

If we are given the equation of a linear relationship, like $y=3x+5$`y`=3`x`+5, then to sketch it we need two points. We can pick any two points we like.

Start by picking any two $x$`x`-values you like, often the $x$`x`-value of $0$0 is a good one to pick because the calculation for y can be quite simple. For our example, $y=3x+5$`y`=3`x`+5 becomes $y=0+5$`y`=0+5, $y=5$`y`=5. This gives us the point $\left(0,5\right)$(0,5)

Similarly look for other easy values to calculate such as $1$1, $10$10, $2$2. I'll pick $x=1$`x`=1. Then for $y=3x+5$`y`=3`x`+5, we have $y=3\times1+5$`y`=3×1+5, $y=8$`y`=8.This gives us the point $\left(1,8\right)$(1,8)

Now we plot the two points and create a line.

The general form of a line is great for identifying both the x and y intercepts easily.

For example, the line $3y+2x-6=0$3`y`+2`x`−6=0

The x intercept happens when the $y$y value is $0$0. $3y+2x-6=0$3y+2x−6=0 $0+2x-6=0$0+2x−6=0 $2x=6$2x=6 $x=3$x=3 |
The y intercept happens when the $x$x value is $0$0. $3y+2x-6=0$3y+2x−6=0 $3y+0-6=0$3y+0−6=0 $3y=6$3y=6 $y=2$y=2 |

From here it is pretty easy to sketch, we find the $x$`x` intercept $3$3, and the $y$`y` intercept $2$2, and draw the line through both.

Start by plotting the single point that you are given.

Remembering that gradient is a measure of change in the rise per change in run, we can step out one measure of the gradient from the original point given.

For a gradient of $4$4 $1$1 unit across and $4$4 units up. | For a gradient of $-3$−3 $1$1 unit across and $3$3 units down. | For a gradient of $\frac{1}{2}$12 $1$1 unit across and $\frac{1}{2}$12 unit up. |

The point can be any point $\left(x,y\right)$(`x`,`y`), or it could be an intercept. Either way, **plot **the point, **step **out the gradient and **draw **your line!

For example, plot the line with gradient $-2$−2 and has $y$`y` intercept of $4$4.

Start with the point, ($y$y intercept of $4$4) |
Step out the gradient, (-$2$2 means $2$2 units down) |
Draw the line |

To sketch linear graphs, it's easiest to substitute in values to find coordinates to put it in gradient-intercept form.

The Gradient-Intercept Form

$y=mx+b$`y`=`m``x`+`b`

where $m$`m` is the gradient and $b$`b` is the $y$`y`-intercept

Our graphs may not always be in this form so we may need to rearrange the equation to make $y$`y` the subject (that means $y$`y` is on one side of the equation and everything else is on the other side).

Sometimes, it doesn't matter. We can sketch a straight line on a graph just by knowing a couple of its features such as a point that lies on the line and it's gradient. At other times, we may need to generate an equation before we sketch it. So other than the gradient-intercept form, we can use:

- Gradient-point formula: $y-y_1=m\left(x-x_1\right)$
`y`−`y`1=`m`(`x`−`x`1)

- Two point formula: $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$
`y`−`y`1`x`−`x`1=`y`2−`y`1`x`2−`x`1

Ok let's look at this in action with some examples.

Plot the graph of the line whose gradient is $-3$−3 and passes through the point $\left(-2,4\right)$(−2,4).

- Loading Graph...

Consider the linear equation $y=3x+1$`y`=3`x`+1.

State the $y$

`y`-value of the $y$`y`-intercept of this line.Using the point $Y$

`Y`as the $y$`y`-intercept, sketch a graph of the equation $y=3x+1$`y`=3`x`+1.Loading Graph...

Graph the linear equation $-6x+3y+24=0$−6`x`+3`y`+24=0 by finding any two points on the line.

- Loading Graph...

On horizontal lines, the $y$`y` value is always the same for every point on the line.

On vertical lines, the $x$`x` value is always the same for every point on the line.

Draw a graph of the line $y=-3$`y`=−3.

- Loading Graph...