The line and its cartesian equation are related in a very special way. Think of the line as an infinite collection of points, all of which lie in a straight line. Each point is uniquely addressed by cartesian coordinates.
The critical thing to remember is that the coordinates of every single point that is on the line, when substituted into the lines equation, will make that equation true. In other words, points on the line will satisfy the equation. Points not on the line will not satisfy the equation.
Worked example
Example 1
Determine whether the point $\left(7,-3\right)$(7,−3) lies on the line $2x+3y=5$2x+3y=5.
Think: Substitute the coordinates of the point into the line equation. If both sides agree, then the point lies on the line–otherwise the point does not lie on the line.
Do:
| LHS | $=$= | $2\times\left(7\right)+3\times\left(-3\right)$2×(7)+3×(−3) |
| $=$= | $14-9$14−9 | |
| $=$= | $5$5 | |
| RHS | $=$= | $5$5 |
Therefore, the point $\left(7,-3\right)$(7,−3) is on this line, as the coordinates when substituted into the equation creates a true statement, where both sides equal $5$5, which means this point satisfies the equation.
Reflect: Now consider the same equation and the point given by $\left(5,0\right)$(5,0). Substitute these coordinates of the point into the line equation.
| LHS | $=$= | $2\times\left(5\right)+3\times\left(0\right)$2×(5)+3×(0) |
| $=$= | $10+0$10+0 | |
| $=$= | $10$10 | |
| RHS | $=$= | $5$5 |
Therefore the point $\left(5,0\right)$(5,0) is not on the line, as the coordinates when substituted into the equation creates a false statement, as $10\ne5$10≠5, which means this point does not satisfy the equation.
Here is the line and the two points tested.
Note that a third point $\left(1,1\right)$(1,1) satisfies the equation, with $2\times\left(1\right)+3\times\left(1\right)=5$2×(1)+3×(1)=5.
This point $\left(1,1\right)$(1,1) , as well as the previously tested point $\left(7,-3\right)$(7,−3), have both been proven to be two points on the same line. Therefore a line can be drawn through both points to represent the equation $2x+3y=5$2x+3y=5.
When any two points can be found that both satisfy the same linear equation, a line can be sketched by simply placing a ruler onto these points and drawing the line. There is only one line that can be drawn through two given points.
Sketching using the $x$x- and $y$y-intercepts
Another way to sketch a linear function is by finding the $x$x-intercept and $y$y-intercept.
The $y$y-coordinate of the $x$x-intercept is always $0$0. Therefore to find the coordinates of the $x$x-intercept, substitute $y=0$y=0 into the equation and solve for $x$x.
Similarly, the $x$x-coordinate of the $y$y-intercept is always $0$0. To find the coordinates of the $y$y-intercept, simply substitute $x=0$x=0 into the equation and solve for $y$y.
In cases where the $x$x- and $y$y-intercepts are equal (usually at the origin $\left(0,0\right)$(0,0)) choose another point, say $x=1$x=1, and substitute $x=1$x=1 into the equation and solve for $y$y to find a second point that lies on that line. Then join the lines by using a rule to draw a straight line between them.
Equations of lines come in all forms.
- The general form is usually written as $Ax+By=C$Ax+By=C but can be written with the constant term ($C$C) on the other side of the equation.
- The point/slope form $y=a+bx$y=a+bx where the slope of the line is given by the coefficient $b$b and the $y$y-intercept given by $a$a.
In all forms, we can use the strategy of setting $x=0$x=0 and then $y=0$y=0 to reveal the two intercepts.
Worked example
Example 2
Find the $x$x-intercept and $y$y-intercept of the equation $3x-y=6$3x−y=6.
Think: We can substitute $x=0$x=0 to find the $y$y-intercept, and then substitute $y=0$y=0 to find the $x$x-intercept.
Do: Let's substitute $x=0$x=0.
| $3x-y$3x−y | $=$= | $6$6 |
| $3\times\left(0\right)-y$3×(0)−y | $=$= | $6$6 |
| $-y$−y | $=$= | $6$6 |
| $\frac{-y}{-1}$−y−1 | $=$= | $\frac{6}{-1}$6−1 |
| $y$y | $=$= | $-6$−6 |
Therefore the coordinates of the $y$y-intercept are $\left(0,-6\right)$(0,−6).
Then we substitute $y=0$y=0 into the equation to find that the $x$x-intercept is $\left(2,0\right)$(2,0).
Practice questions
Question 1
Consider the equation $12x-4y=48$12x−4y=48.
Find the $y$y-value of the $y$y-intercept of the line.
Find the $x$x-value of the $x$x-intercept of the line.
Sketch a graph of the line below.
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Question 2
Sketch a graph of the linear equation $y=3+3x$y=3+3x by determining any two points on the line.
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