A line perpendicular to another line or to a tangent line is o...

Question

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.
y=3x−4; P(1, −1)

y=3x−4; P(1, −1)

Accepted Answer

Welcome back, everyone. Determine the equation of the line perpendicular to the tangent line of the curve Y equals 2 x + 5 at the point Q with X of 2 and Y 9 we're given 4 answer choices. A says Y equals 1/2 x plus 10, B 1/2 x + 8, C-1/2 x + 10, and D 1/2 x + 8. So first of all, let's understand the scenario. Generally, when we have a curve and we're looking for a tangent line at a specific point, we can identify the slope of the tangent line very easily by identifying the derivative of the curve, right, and then substituting for the X coordinate of interest. So we can say that the coordinate of interest is X not, right? And now in this context, when we understand the tangent line, we also want to find a line that is perpendicular to the tangent line, right? So it has that same point, but it basically makes a right angle relative to the tangent line. So our first step is to identify the slope of the tangent line. Soon we will understand why we're doing this. Let's look at the original curve. It is actually a line Y equals 2 x + 5, and we want to find its derivative Y equals 2. Because the derivative of 5 is 02 X gives us 2 multiplied by the derivative of X, and the derivative of X is 1. So why do we need this? Well, Y equals 2, and this means that the slope of the tangent line. It is always 2 regardless what the X is, right? It's independent of X. So this is the slope, right, because the derivative is defined by the slope of the tangent line at a specific point. So we can see that our tangent line. Y of tangent line is simply equal to 2 x plus some coefficient b. We're not worried about the intercept. So we have our slope, right? Our slope, let's call it M. It is equal to 2. Step number 2 is to identify the slope of the normal line, the line that is perpendicular to the tangent line. Let's recall that if two lines are perpendicular to each other, then the product of their slopes M1 multiplied by M2, let's call the original slope of the tangent line M1. This product must be equal to -1, which means that M2, which is going to be the slope of the normal line. Must be equal to -1 divided by m1 and this gives us -1 divided by 2 because that's the slope of the tangent line. Well done. Now, step number 3. What we want to do is write the equation of the line and do that we can use the formula, right, the point slope form, which tells us that if we take Y minus Y1, this must be equal to M multiplied by X minus X1. In this case, because we're looking for the slope of the normal line, we're going to use M2, right? So why remains why. And now we need a point of interest. Now, let's remember that there is a common point between the two lines, which is the point of tangency 2 and 9, which means that Y1 is 9. We're substituting for M2, which is -12, and multiplying that by x minus X1, and X1 is again is the X coordinate of the point of tangency. And now step number 4. It's to simply Take this expression, distribute the terms, and write a simple form, right, with a slope and the intercept. So we get the Y equals distributing the terms, we get -1 X. Minus 1/2 multiplied by -2 gives us 1, so we're adding 1, and then we're adding 9 to both sides. So Y equals -1/2 X + 1 + 9, this gives us 10. So we can see that the correct answer choice to this problem corresponds to the answer choice C. Thank you for watching.

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.y=3x−4; P(1, −1)

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A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.
y=3x−4; P(1, −1)