Graphing Straight Lines

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. 

As an example, consider the line $2x+3y-5=0$2x+3y−5=0 and the point given by $\left(7,-3\right)$(7,−3). If we substitute the coordinates of the point into the line, we see that $2\times\left(7\right)+3\times\left(-3\right)-5=0$2×(7)+3×(−3)−5=0 and so immediately we know the point $\left(7,-3\right)$(7,−3) is on this line. 

We also know that the point $\left(5,0\right)$(5,0)  is not on the line because, when we substitute we find that $2\times\left(5\right)+3\times\left(0\right)-5=5$2×(5)+3×(0)−5=5. Here is the line and the two points we 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=0$2×(1)+3×(1)−5=0, and this leads us to a way to sketch the line. 

If we can find any two points that satisfy the equation, then we can sketch the line by simply placing a ruler onto these points and drawing the line. There is only one line that can be drawn through two points.

By far the easiest way to find two points is to substitute $x=0$x=0 into the line's equation and see what $y$y must be to make the equation true. Then for another point reverse the procedure - substitute $y=0$y=0 into the line's equation and see what $x$x must be to make the equation true. 

Consider the line whose equation is $3x+4y=24$3x+4y=24. If we substitute $x=0$x=0 we find that $0+4y=24$0+4y=24 and so its clear that y must be $6$6. If we then substitute $x=0$x=0 we find that $3x+0=24$3x+0=24 and so its clear that $x$x must be $8$8. Hence we know that $\left(0,6\right)$(0,6) and $\left(8,0\right)$(8,0) are on the line. By using this method the points found are where the line crosses the coordinate axes.  On a graph, we can plot these axes intercepts and then simply draw a line through them.    

Equations of lines come in all sorts of forms. The general form is usually written as $ax+by+c=0$ax+by+c=0 but can be written with the constant term on the other side of the equation. The point/slope form $y=mx+b$y=mx+b where the gradient of the line is given by the coefficient $m$m and the y-intercept given by $b$b. There is also the intercept form $\frac{x}{h}+\frac{y}{k}=1$xh​+yk​=1, called so because the $x$x and $y$y intercepts are $h$h and $k$k respectively. 

In all three forms we can use the strategy of setting $x=0$x=0 and then $y=0$y=0 to reveal the two intercepts.

Consider the example of the line with equation  $3x-y-6=0$3x−y−6=0. We could do one of three things:

• By adding $6$6 to both sides we reveal the alternative general form $3x-y=6$3x−y=6
• By adding $y$y to both sides  we reveal the point/slope form $y=3x-6$y=3x−6.
• By adding $6$6 to both sides and then dividing both sides by 6 we reveal the intercept form $\frac{x}{2}+\frac{y}{-6}=1$x2​+y−6​=1. Carefully note that this form requires a plus sign between the two left hand side terms.

Now we set $x=0$x=0 in any of the forms, and determine that y must be $-6$−6. Then we set $y=0$y=0 in any of the forms, and determine that $x$x must be $2$2. 

Worked examples

Question 1

Question 2

Question 3