Lesson Video: Equations of Straight Lines Mathematics

Video Transcript

In this video, we will learn how to find the equation of a straight line in the form 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one or 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero, given the slope and one point or two different points that it passes through. An equation of a straight line in two variables is any polynomial equation of degree one containing exactly two variables, which are usually 𝑥 and 𝑦. We have already encountered equations of straight lines in the slope–intercept form, 𝑦 equals 𝑚𝑥 plus 𝑐. Here, 𝑚 is the slope of the line and 𝑐 is the value of the 𝑦-intercept.

In this video, we will introduce two new forms in which the equation of a straight line can be given and learn how to derive these two equations given two pieces of data: either the coordinates of two points that lie on a line or the coordinates of a single point and the line’s slope.

Let’s consider first the point–slope form of the equation of a straight line, 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. Here, 𝑚 is the slope of the line and 𝑥 one, 𝑦 one are the coordinates of any given point that lies on the line. One way to think about this form of the equation of a straight line is as a reformulation of the definition of the slope. The slope of a straight line is defined as the change in 𝑦 divided by the change in 𝑥. If we take 𝑥, 𝑦 to be the general point on the line and 𝑥 one, 𝑦 one to be a specific point whose coordinates we know, then we can write the slope 𝑚 as 𝑦 minus 𝑦 one over 𝑥 minus 𝑥 one. Multiplying both sides of this equation by 𝑥 minus 𝑥 one and then swapping the two sides of the equation around gives the point–slope form of the equation of a straight line: 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one.

If we know the value of 𝑚 and the coordinates of one point on the line, we can then substitute these values into this formula to find the equation of any given line. If instead we know the coordinates of two points on the line, we can first calculate the value of 𝑚 by applying the slope formula and then substitute this and the coordinates of either point to find the equation of the line.

Let’s now consider an example in which we find the point–slope form of the equation of a straight line, given its slope and the coordinates of a point on the line.

Find in point–slope form the equation of the graph with slope four that passes through the point two, negative three.

We’re asked to find the equation of this straight line in point–slope form, which we recall is 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one, where 𝑚 is the slope of the line and 𝑥 one, 𝑦 one are the coordinates of any point on the line. We’re told that the slope of the line is four — so this is the value of 𝑚 — and that the line passes through the point two, negative three, so these are the coordinates 𝑥 one, 𝑦 one. Substituting these values gives 𝑦 minus negative three is equal to four multiplied by 𝑥 minus two.

We can simplify the left-hand side of the equation as 𝑦 minus negative three is equal to 𝑦 plus three. But as we’ve been asked to give the answer in point–slope form, we won’t rearrange the equation any further. Hence, we’ve found that the equation of the graph with slope four that passes through the point two, negative three, in point–slope form, is 𝑦 plus three equals four 𝑥 minus two.

Both the point–slope and slope–intercept forms are useful forms of the equation of a straight line and immediately reveal properties such as the slope or 𝑦-intercept of a straight line. But there is a major drawback of both of these forms, which is that not every equation of a straight line can be written in these forms. Specifically, consider lines that are parallel to the 𝑦-axis, which all have equations of the form 𝑥 equals some constant 𝑎. It is not possible to rearrange either of the forms we’ve discussed so far to include this category of straight lines. And so, we introduce a third form, the general form of the equation of a straight line: 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants. Note that the value of 𝑐 here is not the same as the value of 𝑐 in the slope–intercept form.

The general form has the significant advantage that the equation of every straight line, whether it is vertical, horizontal, or diagonal, can be written in this form. It is also straightforward to convert between the general and slope–intercept forms. This can be achieved by subtracting both 𝑎𝑥 and 𝑐 from both sides of the equation of a line given in general form and then dividing both sides by 𝑏, provided 𝑏 is nonzero. The slope and 𝑦-intercept can then be easily identified from this rearranged form.

Let’s now consider an example in which we find the equation of a line in the general form given the coordinates of two points that lie on a line.

A line 𝐿 passes through the points three, three and negative one, zero. Work out the equation of the line, giving your answer in the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero.

We’ve been given the coordinates of two points that lie on line 𝐿 and asked to find its equation in general form. We’ll begin by calculating the slope of the line, using the formula 𝑚 equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one, or change in 𝑦 over change in 𝑥. Substituting the coordinates of the two points gives 𝑚 equals zero minus three over negative one minus three, which simplifies to negative three over negative four and then three-quarters.

We can then recall that if we know the slope of a line and the coordinates of any point on it, we can find its equation by first using the point–slope form of the equation of a straight line: 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. We can use either of the two points and the choice is entirely arbitrary. Substituting 𝑚 equals three-quarters and both 𝑥 one and 𝑦 one equal three gives 𝑦 minus three equals three-quarters of 𝑥 minus three.

Now we need to rearrange this equation into general form. Multiplying both sides by four to eliminate the fractions gives four 𝑦 minus 12 equals three multiplied by 𝑥 minus three. Next, expanding the brackets on the right-hand side gives four 𝑦 minus 12 equals three 𝑥 minus nine. We now need to collect all the terms on the same side of the equation. So subtracting three 𝑥 and adding nine to both sides gives negative three 𝑥 plus four 𝑦 minus three is equal to zero. Equally, we could have collected all the terms on the other side of the equation, which would have resulted in the exact negative of this expression. But as it is equal to zero, the equation is equivalent.

It’s also worth pointing out that if we had chosen to substitute the coordinates of the other given point, negative one, zero, we would have arrived at the same answer once the equation was fully rearranged. We’ve found the equation of the line that passes through the points three, three and negative one, zero in general form is negative three 𝑥 plus four 𝑦 minus three is equal to zero.

In the previous example, we multiplied the entire equation through by four, so that the coefficients in our final answer were all integers. This is generally what we do when given the equation of a straight line in general form. Whilst not strictly essential, it is usual for 𝑎, 𝑏, and 𝑐 to be the smallest possible integers. We eliminate any fractions by multiplying by the denominators. And then if the coefficients all share a common factor, we can divide the entire equation through by this common factor to give the equation in its simplest form. Our answer isn’t incorrect if we don’t do this, but as there are infinitely many alternative equivalent equations, it is usual to give the final answer in this form.

We’ll now consider some other examples, in which we’ll apply the skills we’ve developed to more complex problems involving the equation of a straight line. In the first of these, we’ll find the equation of a straight line given its slope and the fact that it shares its 𝑥-intercept with another straight line.

The line 𝑦 equals four 𝑥 minus eight crosses the 𝑥-axis at the point 𝐴. Find the equation of the line with a slope of negative three passing through the point 𝐴.

We’re given the equation of a straight line in slope–intercept form and told that it crosses the 𝑥-axis at point 𝐴. We can deduce that this line has a slope of four and a 𝑦-intercept of negative eight, although this isn’t entirely necessary. To find the coordinates of point 𝐴, we recall that the 𝑦-coordinate of any point on the 𝑥-axis is zero. So, substituting 𝑦 equals zero into the equation of this straight line gives an equation we can solve to find the 𝑥-coordinate of point 𝐴. Adding eight to both sides of the equation and then dividing by four gives 𝑥 equals two.

At this point, our sketch gives some reassurance that our working so far is correct as the 𝑥-intercept occurs at a positive 𝑥-value. We now know that the line whose equation we want to determine has a slope of negative three and passes through the point 𝐴, which has coordinates two, zero. We can find this equation by substituting these values into the point–slope form of the equation of a straight line: 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. Doing so gives 𝑦 minus zero equals negative three 𝑥 minus two. We can simplify the left-hand side of the equation to simply 𝑦. And as we’re not asked to give the answer in a particular form, we can stop here and leave our answer in point–slope form. The equation of the given straight line is 𝑦 equals negative three multiplied by 𝑥 minus two.

Equally, we could expand the brackets and give our answer in slope–intercept form as 𝑦 equals negative three 𝑥 plus six. Or we could rearrange the equation to the general form and give our answer as three 𝑥 plus 𝑦 minus six is equal to zero. All of these forms are equivalent and correct.

In our next example, we’ll find the equation of a line given the coordinates of one point that lies on it and the knowledge that the line passes through the intersection point of two other lines.

The lines 𝑦 equals negative one-third 𝑥 plus six and 𝑦 equals 𝑥 minus two intersect at the point 𝑆. The point 𝑇 has coordinates negative five, eight. Find the equation of the line passing through the points 𝑆 and 𝑇.

We’re given the coordinates of one point on the line, point 𝑇, and told that the line also passes through point 𝑆, which is the point of intersection of two other straight lines. To find the coordinates of point 𝑆, we need to solve the equations of the two straight lines simultaneously. As the two equations are both of the form 𝑦 equals some function of 𝑥, we can eliminate the 𝑦-variable by equating the two expressions on the other side of the equation to give 𝑥 minus two equals negative one-third 𝑥 plus six. Multiplying both sides of this equation by three gives three 𝑥 minus six equals negative 𝑥 plus 18. And then collecting like terms by adding 𝑥 and six to both sides gives four 𝑥 equals 24. Finally, we can divide both sides of the equation by four to find that the 𝑥-coordinate of point 𝑆 is six.

To find the 𝑦-coordinate, we can substitute 𝑥 equals six into either equation. But it will be simpler to use the equation that doesn’t involve fractions. Substituting 𝑥 equals six into the equation 𝑦 equals 𝑥 minus two, we find that the 𝑦-coordinate of point 𝑆 is four.

So, we now know the coordinates of both points that lie on the line whose equation we wish to find. We can find the slope of the line using the formula 𝑚 equals change in 𝑦 over change in 𝑥, or 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. Substituting the coordinates of 𝑆 and 𝑇 gives 𝑚 equals eight minus four over negative five minus six, which is four over negative 11 or negative four elevenths. We can then find the equation of the straight line using the point–slope form: 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. We can substitute the coordinates of either point 𝑆 or 𝑇 for 𝑥 one, 𝑦 one. But it probably makes more sense to use point 𝑇 as we were given these coordinates explicitly. We obtain 𝑦 minus eight equals negative four elevenths 𝑥 minus negative five.

We’re not asked to give our answer in a particular form, so we can leave it as it is. The only simplification we’ll make is to rewrite 𝑥 minus negative five as 𝑥 plus five. We’ve found that the equation of the line passing through points 𝑆 and 𝑇 is 𝑦 minus eight equals negative four elevenths 𝑥 plus five.

Let’s finish by summarizing the key points from this video. Given the coordinates of any two points that lie on a straight line, 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, we can find the slope of that line using the formula change in 𝑦 over change in 𝑥, or 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. We recalled the slope–intercept form of the equation of a straight line, 𝑦 equals 𝑚𝑥 plus 𝑐, where 𝑚 represents the slope and 𝑐 represents the 𝑦-intercept of the line. We then introduced a new form of the equation of a straight line, the point–slope form, 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one.

Given the slope 𝑚 of a line and the coordinates 𝑥 one, 𝑦 one of any point that lies on that line, we can find the equation of the line by substituting 𝑚, 𝑥 one, and 𝑦 one into the point–slope form of the equation of a straight line. We then saw that whilst the first two forms of the equation of a straight line are useful in certain contexts, they both have the major drawback that they do not include all straight lines. And in particular, the equations of vertical lines cannot be expressed in either of these forms. We therefore introduced the general form of the equation of a straight line: 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero, which has the significant advantage that the equation of any straight line can be written in this form.

Finally, for those lines whose equations can be written in multiple forms, we can convert between these different forms by rearranging the equation.