MHB How does reflecting points in a line affect the equation of the original line?

[mathdad]

Pick any three points on the line y = (x/2) + 3.

1. Reflect each point in the line y = x.

Do I simply exchange the x and y coordinates of the chosen 3 points?

2. Show that the 3 reflected points all lie on one line.

I need one or two hints here.

3. What is the equation of the line for part (2)?

 

[topsquark]

Pick any three points on the line y = (x/2) + 3.

1. Reflect each point in the line y = x.

Do I simply exchange the x and y coordinates of the chosen 3 points?

Yes.

2. Show that the 3 reflected points all lie on one line.

I need one or two hints here.

How do you know that points are on a line? Easy: All points have to satisfy the line equation y = mx + b for some m and b. Start with any two of the reflected points and find the equation of the line between them. If all three reflected points are on the line then the third point will also satisfy the same line equation.

-Dan

 

[MarkFL]

Another way to show 3 points are collinear is to pick two distinct pairs from the 3 points, and show that the slope between both pairs is the same. :D

 

[Daviddur]

It seems to me that you can graph on 2 point. That it will look like this Line y = (x/2) + 3.

View attachment 6426

 

[Evgeny.Makarov]

It seems to me that you can graph on 2 point. That it will look like this Line y = (x/2) + 3.

The scale along the $x$ line is not clear from the picture, so it is impossible to say whether this is the graph of $y=x/2+3$.

 

[HallsofIvy]

Reflecting a point (a, b) about the line y= x gives the point (b, a). That is, it swaps the two coordinates. Given that y= (x/2)+ 3, x/2= y- 3 and than x= 2(y- 3)= 2y- 6. Swapping x and y, y= 2x- 6.

Another way of thinking about it: the function y= (x/2)+ 3 says "first divide x by 2 then add 3". Reflecting about the line y= x gives the inverse function, which is just doing the reverse. The reverse of "divide by 2" is "multiply by 2" and the reverse of "add 3" is "subtract 3". Further, reversing the function reverses the order in which those are done. So the reverse of "first divide by 2 and the add 3" is "first subtract three and then multiply by 2": y= 2(x- 3)= 2x- 6 as before.

 

[mathdad]

Reflecting a point (a, b) about the line y= x gives the point (b, a). That is, it swaps the two coordinates. Given that y= (x/2)+ 3, x/2= y- 3 and than x= 2(y- 3)= 2y- 6. Swapping x and y, y= 2x- 6.

Another way of thinking about it: the function y= (x/2)+ 3 says "first divide x by 2 then add 3". Reflecting about the line y= x gives the inverse function, which is just doing the reverse. The reverse of "divide by 2" is "multiply by 2" and the reverse of "add 3" is "subtract 3". Further, reversing the function reverses the order in which those are done. So the reverse of "first divide by 2 and the add 3" is "first subtract three and then multiply by 2": y= 2(x- 3)= 2x- 6 as before.

Very informative.