Prove Function Reflection: Mirrored in y=x Line

[eleventhxhour]

If (a, b) = f, then (b, a)= f^-1. The claim has been made that the graph of the inverse of a function is the mirror image of the function, reflected in the line y = x. To prove this you must

a) Show that the line that passes through the points (a, b) and (b, a) intersect the line y = x at a right angle.

b) Show that the distances between each point and the line y = x are the same. (To do this you MIGHT have to determine the point of intersection of the line passing though (a, b) and (b, a) and the line y = x).

Please help! I know that you can't really graph it on here, but it'd be so much help if you explained what the question means and how to solve it. I really don't understand it, so any sort of help would be really appreciated. Thanks!

 

[Evgeny.Makarov]

Here is a picture.

View attachment 2200

 

[MarkFL]

Let's start with part a). To show that two lines are perpendicular, you can simply show that the product of their slopes is -1, as demonstrated here:

http://mathhelpboards.com/math-notes-49/perpendicular-lines-product-their-slopes-2953.html

So, what is the slope of the line passing through the points $(a,b)$ and $(b,a)$? What is the slope of the line $y=x$? What is the product of the two slopes?

 

[eleventhxhour]

Let's start with part a). To show that two lines are perpendicular, you can simply show that the product of their slopes is -1, as demonstrated here:

http://mathhelpboards.com/math-notes-49/perpendicular-lines-product-their-slopes-2953.html

So, what is the slope of the line passing through the points $(a,b)$ and $(b,a)$? What is the slope of the line $y=x$? What is the product of the two slopes?

Would the slope of the line passing through the points $(a,b)$ and $(b,a)$ be $-1$? And the slope of the line $y=x$ is $1$? The product is $-1$.

I understand how the line intersects y=x at a right angle, but I still don't know how to show that the distance is the same.

Also, there was another very small part of this question that I didn't understand: "Does this prove that the claim is true for any function? Why?"

 

[MarkFL]

Yes, you are correct concerning the slopes and their product. (Yes)

Next, find the point where the two lines intersect, and then apply the distance formula. Or, simply apply this formula:

http://mathhelpboards.com/math-notes-49/finding-distance-between-point-line-2952.html

 

[eleventhxhour]

Yes, you are correct concerning the slopes and their product. (Yes)

Next, find the point where the two lines intersect, and then apply the distance formula. Or, simply apply this formula:

http://mathhelpboards.com/math-notes-49/finding-distance-between-point-line-2952.html

I'm not really sure how you'd find the point where the two lines intersect. The points are (a, b) and (b, a), but the values aren't given for them. Would you just assume two random points? And then would you find the equation of that line?

I may be completely wrong haha.

 

[MarkFL]

Well, you know the slope is -1 and you know two points on the line (all you need to know is one) and so apply the point-slope formula to get the line:

\(\displaystyle y-b=-1(x-a)\)

\(\displaystyle y=-x+a+b\)

So, to find where this line intersects with the line $y=x$ solve:

\(\displaystyle x=-x+a+b\)

for $x$ and your point of intersection is $(x,x)$. Then apply the distance formula separately to the points $(a,b),\,(b,a)$ and the point of intersection you found. That is, show that the distance between $(a,b)$ and $(x,x)$ is the same as the distance between $(b,a)$ and $(x,x)$.

 

[eleventhxhour]

Well, you know the slope is -1 and you know two points on the line (all you need to know is one) and so apply the point-slope formula to get the line:

\(\displaystyle y-b=-1(x-a)\)

\(\displaystyle y=-x+a+b\)

So, to find where this line intersects with the line $y=x$ solve:

\(\displaystyle x=-x+a+b\)

for $x$ and your point of intersection is $(x,x)$. Then apply the distance formula separately to the points $(a,b),\,(b,a)$ and the point of intersection you found. That is, show that the distance between $(a,b)$ and $(x,x)$ is the same as the distance between $(b,a)$ and $(x,x)$.

So, doing the distance formula, would you get this:
$$d = \sqrt(a-x)^2 + (b-y)^2$$

And then I guess you'd solve it? That would get a lot of different variables though. It seems pretty complicated for a grade 11 question.

 

[MarkFL]

You want to find the point of intersection first, so solving the equation I gave you for $x$, we get:

\(\displaystyle x=\frac{a+b}{2}\)

And so the point of intersection is:

\(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\)

Next, we check the two distances to see that they are equal:

1.) Distance between \(\displaystyle (a,b)\) and \(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\):

\(\displaystyle d_1=\sqrt{\left(a-\frac{a+b}{2} \right)^2+\left(b-\frac{a+b}{2} \right)^2}\)

2.) Distance between \(\displaystyle (b,a)\) and \(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\):

\(\displaystyle d_2=\sqrt{\left(b-\frac{a+b}{2} \right)^2+\left(a-\frac{a+b}{2} \right)^2}\)

We can see without simplifying that by the commutative law of addition, the two distances are in fact the same.

 

[eleventhxhour]

You want to find the point of intersection first, so solving the equation I gave you for $x$, we get:

\(\displaystyle x=\frac{a+b}{2}\)

And so the point of intersection is:

\(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\)

Next, we check the two distances to see that they are equal:

1.) Distance between \(\displaystyle (a,b)\) and \(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\):

\(\displaystyle d_1=\sqrt{\left(a-\frac{a+b}{2} \right)^2+\left(b-\frac{a+b}{2} \right)^2}\)

2.) Distance between \(\displaystyle (b,a)\) and \(\displaystyle \left(\frac{a+b}{2},\frac{a+b}{2} \right)\):

\(\displaystyle d_2=\sqrt{\left(b-\frac{a+b}{2} \right)^2+\left(a-\frac{a+b}{2} \right)^2}\)

We can see without simplifying that by the commutative law of addition, the two distances are in fact the same.

Thanks so much! That helps a lot. :)

Also, this was part b) of the question, which I'm not entirely sure how to do/explain:

"Does doing a) prove that the claim is true for any function? Why?"

 

[MarkFL]

Do you mean that is part c) of the question? Part b) was about the distances...how is the question actually structured?

 

[eleventhxhour]

Do you mean that is part c) of the question? Part b) was about the distances...how is the question actually structured?

Oh, my bad! Sorry about that. It's pretty much part b) to that entire set of questions, but on the actual sheet it's #2. I was just trying to make it simpler/easier to read. Here are the two questions exactly as written on the sheet:

1. If $(a, b) = f$, then $(b, a)= f^-1$. The claim has been made that the graph of the inverse of a function is the mirror image of the function, reflected in the line y = x. To prove this you must

a) Show that the line that passes through the points $(a, b)$ and $(b, a)$ intersect the line $y = x$ at a right angle.

b) Show that the distances between each point and the line $y = x $are the same. (To do this you MIGHT have to determine the point of intersection of the line passing though $(a, b)$ and $(b, a)$ and the line $y = x$).

2. Does doing 1. prove that this is true for any function? Why?

I'd assume that when it says "doing 1", it means proving that both a) and b) are true for any function.

Sorry for the confusion! And thanks again for your help.

 

[MarkFL]

Do all functions have an inverse function?

 

[eleventhxhour]

Do all functions have an inverse function?

Well, they may have an inverse that is not a function.

 

[MarkFL]

That notwithstanding, do you find any point $(a,b)$ for which part 1.) does not apply?

 

[eleventhxhour]

That notwithstanding, do you find any point $(a,b)$ for which part 1.) does not apply?

Hm, no. You can sub in any point for $(a,b)$ and it will work.