The points in this table lie on a line. Find the equation of the line in slope–intercept form.
Let’s begin by recalling the slope–intercept form of the equation of a straight line. It’s 𝑦 equals 𝑚𝑥 plus 𝑏, where the coefficient of 𝑥, that’s 𝑚, represents the slope of the line and the constant term, that’s 𝑏, represents its 𝑦-intercept.
We’ve been given four pairs of values of corresponding 𝑥- and 𝑦-values that lie on this line. So these are the coordinates of points that lie on the line. For example, when 𝑥 is equal to five, 𝑦 is equal to negative eight. So the point with coordinates five, negative eight is on the line. There are a couple of different ways that we could approach this problem. In our first method, we’re going to use the coordinates of some of these points to set up some linear equations, which we can then solve simultaneously.
For the first point, we know that when 𝑥 is equal to five, 𝑦 is equal to negative eight. So substituting these values for 𝑥 and 𝑦 into the equation 𝑦 equals 𝑚𝑥 plus 𝑏 gives the equation negative eight equals five 𝑚 plus 𝑏. The second pair of values you’ve been given is 10, negative 15. So when 𝑥 is equal to 10, 𝑦 is equal to negative 15. Substituting this pair of values for 𝑥 and 𝑦 gives the equation negative 15 equals 10𝑚 plus 𝑏. And we now have a pair of linear simultaneous equations in the unknowns 𝑚 and 𝑏.
We notice that as the equations have the same coefficient of 𝑏, in each case the coefficient of 𝑏 is simply positive one, then if we subtract one equation from the other, this will eliminate the 𝑏-terms to give an equation in 𝑚 only. Let’s subtract equation one from equation two, because equation two has the larger coefficient of 𝑚. Doing so will eliminate the 𝑏-terms as we already said and gives negative 15 minus negative eight is equal to 10𝑚 minus five 𝑚.
Negative 15 minus negative eight is negative 15 plus eight, which is negative seven. So this simplifies to negative seven equals five 𝑚. To solve for 𝑚, we divide both sides of the equation by five, giving 𝑚 equals negative seven over five. We’ve therefore found the value of one of the unknowns, the slope of this line.
To find the value of 𝑏, we need to substitute the value of 𝑚 we’ve just calculated into either of the two equations. Let’s choose to substitute into equation one. This gives negative eight is equal to five multiplied by negative seven over five plus 𝑏. On the right-hand side, the five in the numerator of the first term will cancel with the five in the denominator. So the equation simplifies to negative eight is equal to negative seven plus 𝑏. We can solve this equation for 𝑏 by adding seven to each side, and it gives negative one is equal to 𝑏 or 𝑏 equals negative one.
We’ve therefore found the values of both the unknowns 𝑚 and 𝑏. All that remains is to substitute these values into the equation 𝑦 equals 𝑚𝑥 plus 𝑏, giving 𝑦 equals negative seven over five 𝑥 minus one. And so we’ve found the equation of this straight line. In our working though, we’ve only used two of the points we were told lie on the line, so we can check our working by using either or both of the other two points.
We know that when 𝑥 is equal to 15, 𝑦 should be equal to negative 22. So let’s check. When 𝑥 is equal to 15, our equation gives 𝑦 is equal to negative seven over five multiplied by 15 minus one. We can cancel a factor of five in the numerator and denominator of the first term to give 𝑦 equals negative seven over one multiplied by three minus one. That gives negative 21 minus one, which is equal to negative 22. This is the correct value for 𝑦, and so this confirms that our equation is correct. We can perform the same check with the final point if we wish. When 𝑥 is equal to 20, we find that 𝑦 is indeed equal to negative 29.
So by forming and solving two linear simultaneous equations, we were able to determine the values of 𝑚 and 𝑏 and hence found that the equation of this line in slope–intercept form is 𝑦 equals negative seven over five 𝑥 minus one.
Let’s consider, though, another method we could have used. We should recall that given the coordinates of two points that lie on a straight line 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the slope of the line is given by the change in 𝑦 over change in 𝑥. That’s 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. We can therefore choose any two of the four points we know lie on the line to calculate its slope.
The choice is entirely up to us, so let’s use the two coordinates in the middle. Substituting the coordinates of these points gives negative 22 minus negative 15 in the numerator and 15 minus 10 in the denominator. Negative 22 minus negative 15 is the same as negative 22 plus 15. So this simplifies to negative seven over five, which we see is the same as the slope of the line we calculated using our first method.
To find the value of 𝑏, the 𝑦-intercept, we should notice that there is a pattern in the values we’ve been given in the table. The 𝑥-values are a sequence of five, 10, 15, 20; they increase by five each time. The 𝑦-values also form a linear sequence. Negative eight, negative 15, negative 22, negative 29 is a linear or arithmetic sequence which decreases by seven each time. The 𝑦-intercept corresponds to an 𝑥-value of zero, which is the 𝑥-value we would get if we went one term back in our sequence, if we went down by five. And so the 𝑦-value will be the next term in the sequence of 𝑦-values if we go the other way.
Now remember we said this sequence was decreasing by seven. So if we go in the opposite direction, then we will find this term by adding seven. Negative eight plus seven is negative one. So we find again that the 𝑦-intercept of the line is negative one. Using two different methods then, we found that the equation of the line on which all of these points lie, in slope–intercept form, is 𝑦 equals negative seven over five 𝑥 minus one.