Point (a, b) Reflected About Two Lines

[mathdad]

This question is found in the challenge section of the textbook. This is a two-part question. Like always, I am looking for steps or hints that will allow me to work out the math.

1. If the point (a, b) is reflected in the line y = 3x, show that the coordinates of the reflected points are as follows:

[(3b - 4a)/5, (3a + 4b)/5]

2. If the point (a, b) is reflected in the line y = mx, what are the coordinates of the reflected point?

 

[topsquark]

This question is found in the challenge section of the textbook. This is a two-part question. Like always, I am looking for steps or hints that will allow me to work out the math.

1. If the point (a, b) is reflected in the line y = 3x, show that the coordinates of the reflected points are as follows:

[(3b - 4a)/5, (3a + 4b)/5]

2. If the point (a, b) is reflected in the line y = mx, what are the coordinates of the reflected point?

What have you tried?

-Dan

 

[MarkFL]

Let's generalize the method I posted http://mathhelpboards.com/pre-calculus-21/reflection-about-y-x-20690-post94058.html#post94058 to reflect a given point $(a,b)$ about the line $y=mx+k$, where ($m\ne0$). We will call the reflected point $(a',b')$.

First, we want the distance $d$ between the given point and the line:

\(\displaystyle d=\frac{|ma+k-b|}{\sqrt{m^2+1}}\)

So, we know $(a'b')$ must lie along the line:

\(\displaystyle y=-\frac{1}{m}(x-a)+b\)

Hence:

\(\displaystyle mb'-mb=a-a'\tag{1}\)

We also know the distance from $(a'b')$ to the line must be $d$, so we have:

\(\displaystyle d=\frac{|ma'+k-b'|}{\sqrt{m^2+1}}\)

Thus, we want to look at (to ensure the two points are distinct):

\(\displaystyle ma+k-b=-ma'-k+b'\)

\(\displaystyle b+b'-k=ma+ma'+k\tag{2}\)

From (1), we find:

\(\displaystyle b'=\frac{mb+a-a'}{m}\)

And from (2):

\(\displaystyle b'=ma+ma'-b+2k\)

Hence:

\(\displaystyle \frac{mb+a-a'}{m}=ma+ma'-b+2k\)

Solving for $a'$, there results:

\(\displaystyle a'=\frac{(1-m^2)a+2m(b-k)}{m^2+1}\)

And so plugging this value into either expression for $b'$ above gives us (after simplification):

\(\displaystyle b'=\frac{(m^2-1)b+2(ma+k)}{m^2+1}\)

Now we have a general formula. (Clapping)

In the given problem, we have $m=3$ and $k=0$, thus:

\(\displaystyle a'=\frac{(1-3^2)a+2(3)(b-0)}{3^2+1}=\frac{-8a+6b}{10}=\frac{3b-4a}{5}\)

\(\displaystyle b'=\frac{(3^2-1)b+2((3)a+0)}{3^2+1}=\frac{8b+6a}{10}=\frac{4b+3a}{5}\)

 

[mathdad]

Let's generalize the method I posted http://mathhelpboards.com/pre-calculus-21/reflection-about-y-x-20690-post94058.html#post94058 to reflect a given point $(a,b)$ about the line $y=mx+k$, where ($m\ne0$). We will call the reflected point $(a',b')$.

First, we want the distance $d$ between the given point and the line:

\(\displaystyle d=\frac{|ma+k-b|}{\sqrt{m^2+1}}\)

So, we know $(a'b')$ must lie along the line:

\(\displaystyle y=-\frac{1}{m}(x-a)+b\)

Hence:

\(\displaystyle mb'-mb=a-a'\tag{1}\)

We also know the distance from $(a'b')$ to the line must be $d$, so we have:

\(\displaystyle d=\frac{|ma'+k-b'|}{\sqrt{m^2+1}}\)

Thus, we want to look at (to ensure the two points are distinct):

\(\displaystyle ma+k-b=-ma'-k+b'\)

\(\displaystyle b+b'-k=ma+ma'+k\tag{2}\)

From (1), we find:

\(\displaystyle b'=\frac{mb+a-a'}{m}\)

And from (2):

\(\displaystyle b'=ma+ma'-b+2k\)

Hence:

\(\displaystyle \frac{mb+a-a'}{m}=ma+ma'-b+2k\)

Solving for $a'$, there results:

\(\displaystyle a'=\frac{(1-m^2)a+2m(b-k)}{m^2+1}\)

And so plugging this value into either expression for $b'$ above gives us (after simplification):

\(\displaystyle b'=\frac{(m^2-1)b+2(ma+k)}{m^2+1}\)

Now we have a general formula. (Clapping)

In the given problem, we have $m=3$ and $k=0$, thus:

\(\displaystyle a'=\frac{(1-3^2)a+2(3)(b-0)}{3^2+1}=\frac{-8a+6b}{10}=\frac{3b-4a}{5}\)

\(\displaystyle b'=\frac{(3^2-1)b+2((3)a+0)}{3^2+1}=\frac{8b+6a}{10}=\frac{4b+3a}{5}\)

Thank you very much. Great work!