### Video Transcript

Find the equation of the normal to the curve 𝑦 equals five 𝑥 plus nine over three 𝑥 minus five at one, negative seven.

Let’s begin by recalling what we know about the normal to a curve. Suppose we have a curve. It doesn’t really matter what it looks like. The tangent to that curve at a particular point has the same slope as the curve itself at that point, whereas the normal to a curve at any given point is perpendicular to the tangent at that same point. What this means is that if the slope of the tangent is some value 𝑚 one, then the slope of the normal will be negative one over 𝑚 one. It is the negative reciprocal of 𝑚 one. Because we know that the slopes of perpendicular lines multiply together to give negative one or are the negative reciprocals of one another.

We’re asked to find the equation of the normal to this curve 𝑦 equals five 𝑥 plus nine over three 𝑥 minus five at a particular point. We can recall that a general method for finding the equation of a straight line is to use the point–slope form of the equation of a straight line. 𝑦 minus 𝑦 one equals 𝑚 multiplied by 𝑥 minus 𝑥 one, where 𝑥 one, 𝑦 one are the coordinates of any point on the line and 𝑚 is its slope. We’ve been given the coordinates of a point we can use for 𝑥 one, 𝑦 one, the point one, negative seven. But we don’t yet know the value of 𝑚.

To find the slope of the normal, we’re first going to need to find the slope of the tangent by recalling that the slope of the tangent is the same as the slope of the curve itself at that point. And we can find the slope of a curve at any point using differentiation. We want to find an expression for d𝑦 by d𝑥, the first derivative of the equation of the curve, which is also its slope function. Now, we notice that the equation 𝑦 is the quotient of two differentiable functions, five 𝑥 plus nine and three 𝑥 minus five. We can therefore find this derivative using the quotient rule.

This tells us that for two differentiable functions of 𝑥, 𝑢, and 𝑣, the derivative with respect to 𝑥 of their quotient, 𝑢 over 𝑣, is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. So we’re going to let 𝑢 equal the function in the numerator, that’s five 𝑥 plus nine, and 𝑣 equal the function in the denominator. That’s three 𝑥 minus five. We can then find each of their derivatives by recalling the power rule of differentiation. The derivative with respect to 𝑥 of five 𝑥 is simply five. And the derivative with respect to 𝑥 of nine or indeed any constant is simply zero. In the same way, the derivative with respect to 𝑥 of three 𝑥 is three and the derivative with respect to 𝑥 of the constant negative five is also zero.

So we have d𝑢 by d𝑥 equals five and d𝑣 by d𝑥 equals three. We can then find our expression for d𝑦 by d𝑥 by substituting into the quotient rule. We first have 𝑣 times d𝑢 by d𝑥. That’s three 𝑥 minus five multiplied by five. Then we subtract 𝑢, that’s five 𝑥 plus nine, multiplied by d𝑣 by d𝑥, which is three. This is all divided by 𝑣 squared. That’s three 𝑥 minus five squared. We can simplify a little by distributing the parentheses in the numerator to give 15𝑥 minus 25 minus 15𝑥 minus 27 all over three 𝑥 minus five all squared. And this then simplifies to negative 52 over three 𝑥 minus five all squared because the 15𝑥 and the negative 15𝑥 in the numerator cancel each other out.

So we found the general slope function for this curve. But we need to evaluate it at a particular point, the point one, negative seven, which means we need to substitute 𝑥 equals one into this formula. Doing so gives d𝑦 by d𝑥 equals negative 52 over three multiplied by one minus five squared. Three multiplied by one is three. And subtracting five gives negative two. So we have negative 52 over negative two squared. That’s negative 52 over four, which is equal to negative 13. Now, we may be tempted to use this value of negative 13 as the slope that we’re going to substitute into the equation of straight line. But remember, this is the slope of the tangent of the point one negative, seven. And we’re asked to find the equation of the normal.

If the slope of the tangent is negative 13, then the slope of the normal is the negative reciprocal of this. It’s negative one over negative 13. And of course, the negatives cancel out, leaving one over 13. We’re now able to substitute the point one, negative seven and the slope one over 13 into the point–slope form of the equation of a straight line to calculate the equation of the normal. Let’s clear some space to do this.

Substituting the point one, negative seven for 𝑥 one, 𝑦 one and one over 13 for the slope 𝑚, we have the equation 𝑦 minus negative seven is equal to one over 13 multiplied by 𝑥 minus one. 𝑦 minus negative seven is, of course, equal to 𝑦 plus seven. And we can then multiply every term in this equation by 13 to eliminate the fraction. This gives 13𝑦 plus 91 is equal to 𝑥 minus one. Finally, we’ll group all of the terms on the left-hand side of the equation by subtracting 𝑥 and adding one to each side. This gives us the equation of the normal to the given curve at the point one, negative seven. It is 13𝑦 minus 𝑥 plus 92 equals zero.