MHB How do I sketch the reflection of line segment AB about the line y = x?

[mathdad]

The endpoints of line AB are given to be (-1, -2) and
(-5, -2). Sketch the reflection of line segment AB about the line y = x.

As I understand this concept, I must show that the line y = x is the perpendicular bisector of the line segment AB.

Two points are given. This tells me to find the slope m.

m = (-2 - (-2))/(-5 - (-1))

m = (-2 + 2)/(-5 + 1)

m = 0/-4

m = 0

When m = 0, we are talking about a horizontal line.

Where do I go from here?

 

[MarkFL]

A reflection about the line $y=x$ simply requires swapping the $x$ and $y$ coordinates.

To see why this is true, consider a point in the plane $\left(x_1,y_1\right)$. Its distance $d$ to the line $y=x$ is given by:

$$d=\frac{\left|x_1-y_1\right|}{\sqrt{2}}$$

So, we want to find the other point $\left(x_1,y_1\right)$ with the same distance, only on the other side of $y=x$, that lies along the line:

$$y=-(x-x_1)+y_1$$

Hence:

$$y_2-y_1=x_1-x_2\tag{1}$$

$$d=\frac{\left|x_2-y_2\right|}{\sqrt{2}}$$

This gives us two cases:

i) $$x_2-y_2=x_1-y_1\implies y_2-y_1=x_2-x_1$$

Adding this to (1), we find:

$$y_1=y_2$$

This means we have the original point, so we discard this.

ii) $$x_2-y_2=y_1-x_1\implies y_2+y_1=x_2+x_1$$

Adding this to (1), we find:

$$y_2=x_1\implies x_2=y_1$$

Thus, we find the reflection of $\left(x_1,y_1\right)$ about the line $y=x$ is the point $\left(y_1,x_1\right)$.

 

[mathdad]

A reflection about the line $y=x$ simply requires swapping the $x$ and $y$ coordinates.

To see why this is true, consider a point in the plane $\left(x_1,y_1\right)$. Its distance $d$ to the line $y=x$ is given by:

$$d=\frac{\left|x_1-y_1\right|}{\sqrt{2}}$$

So, we want to find the other point $\left(x_1,y_1\right)$ with the same distance, only on the other side of $y=x$, that lies along the line:

$$y=-(x-x_1)+y_1$$

Hence:

$$y_2-y_1=x_1-x_2\tag{1}$$

$$d=\frac{\left|x_2-y_2\right|}{\sqrt{2}}$$

This gives us two cases:

i) $$x_2-y_2=x_1-y_1\implies y_2-y_1=x_2-x_1$$

Adding this to (1), we find:

$$y_1=y_2$$

This means we have the original point, so we discard this.

ii) $$x_2-y_2=y_1-x_1\implies y_2+y_1=x_2+x_1$$

Adding this to (1), we find:

$$y_2=x_1\implies x_2=y_1$$

Thus, we find the reflection of $\left(x_1,y_1\right)$ about the line $y=x$ is the point $\left(y_1,x_1\right)$.

Do I switch the original coodinates of the given points, plot the points on the xy-plane and then connect them with a straight line?

 

[MarkFL]

Do I switch the original coodinates of the given points, plot the points on the xy-plane and then connect them with a straight line?

Yes. (Yes)

 

[HallsofIvy]

The endpoints of line AB are given to be (-1, -2) and
(-5, -2). Sketch the reflection of line segment AB about the line y = x.

As I understand this concept, I must show that the line y = x is the perpendicular bisector of the line segment AB.

Well, there's your problem- you understand wrong. You need to find two new points, A' and B', such y= x is the perpendicular bisector of both the line segment from A to A' and the line segment from B to B'.

If A were (x, y) and B were (y, x), that is, if A were something like (2, 3) and B were (3, 2), then y= x would be the perpendicular bisector of AB.

Two points are given. This tells me to find the slope m.

m = (-2 - (-2))/(-5 - (-1))

m = (-2 + 2)/(-5 + 1)

m = 0/-4

m = 0

When m = 0, we are talking about a horizontal line.

Where do I go from here?

 

[mathdad]

Well, there's your problem- you understand wrong. You need to find two new points, A' and B', such y= x is the perpendicular bisector of both the line segment from A to A' and the line segment from B to B'.

I know what to do thanks to Mark.