Find the slope of the straight line whose equation is six 𝑦 equals 𝑎𝑥 plus nine and passes through the point 𝐴 with coordinate seven, negative two.
The slope or gradient of a straight line is the steepness of that line. Lines with a positive gradient slope upwards from left to right, whereas lines with a negative gradient slope downwards from left to right. It’s relatively straightforward to determine the slope of a straight line if its equation is already in what we call slope–intercept form, 𝑦 equals 𝑚𝑥 plus 𝑐. As here, the value 𝑚 represents the slope of the line and the value 𝑐 represents the 𝑦-intercept. That’s the 𝑦-value at which the straight line intersects the 𝑦-axis.
Looking at the equation of the line we’ve been given, we can see that it is nearly in this form, although instead of just being 𝑦 on the left-hand side, it is six 𝑦. However, we can bring it into the form 𝑦 equals 𝑚𝑥 plus 𝑐 if we divide both sides of the equation by six. On the left, dividing by six gives 𝑦. And on the right-hand side, dividing by six gives the expression 𝑎𝑥 plus nine all over six.
We can separate the expression on the right-hand side of this equation into the sum of two separate fractions with a common denominator of six. We can write it as 𝑎𝑥 over six or 𝑎 over six 𝑥 plus nine over six. The value of nine over six can actually be simplified by dividing both the numerator and denominator of this fraction by three, to give the simplified fraction three over two. Now, comparing the equation of our line with the general form, we see that the slope of our line is 𝑎 over six because the coefficient of 𝑥 — that’s the value in front of the 𝑥 — is 𝑎 over six. And although we haven’t been asked to give the 𝑦-intercept of our line, we can see that it is equal to three over two.
In order to determine the slope of this straight line then, we need to work out the value of this constant 𝑎. To do so, we use the fact that this straight line passes through the point capital 𝐴, which has coordinates seven, negative two. As this point lies on our straight line, this pair of 𝑥- and 𝑦-values must satisfy the equation of the straight line. So, we know that when 𝑥 is equal to seven, 𝑦 must be equal to negative two. And we have a pair of values for 𝑥 and 𝑦 which we can substitute into the equation of our line in order to find the value of 𝑎.
Let’s substitute into the equation of our straight line in the form it was originally given in because this form doesn’t involve any fractions. Substituting negative two for 𝑦 and seven for 𝑎, we get the equation six multiplied by negative two is equal to 𝑎 multiplied by seven plus nine. Six multiplied by negative two is equal to negative 12. Remember, a positive multiplied by a negative gives a negative answer. And on the right-hand side, 𝑎 multiplied by seven can be written as seven 𝑎. So, we have the equation negative 12 equals seven 𝑎 plus nine, which we now wish to solve to find the value of 𝑎.
We begin by subtracting nine from each side of the equation. On the right-hand side, we’re just left with seven 𝑎. And on the left-hand side, negative 12 minus a further nine takes us to negative 21. Our final step in solving this equation is to divide both sides by seven. On the right, seven 𝑎 divided by seven just gives 𝑎. And on the left, negative 21 divided by seven gives negative three. Remember, a negative divided by a positive gives a negative answer. So, we found the value of 𝑎. 𝑎 is equal to negative three.
Our final step is to substitute this value for 𝑎 into the expression we found for the slope of our line. Remember, we said that the slope of our line was equal to 𝑎 over six. So, substituting 𝑎 equals negative three gives that the slope of the line is equal to negative three over six. We can simplify this fraction though by dividing both the numerator and denominator by three. Three divided by three is one, and six divided by three is two.
So, we’ve found that the slope of the straight line whose equation is six 𝑦 equals 𝑎𝑥 plus nine and passes through the point with coordinate seven, negative two is negative one-half.