MHB Point (a, b) Reflected About Two Lines

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The discussion focuses on reflecting a point (a, b) across the line y = 3x, resulting in the coordinates [(3b - 4a)/5, (3a + 4b)/5]. The participants explore a general method for reflecting points across lines of the form y = mx + k, deriving formulas for the reflected coordinates. They establish the distance from the point to the line and create equations to find the reflected point's coordinates. The derived formulas are confirmed with the specific case of m = 3 and k = 0, leading to the final results for the reflected point. This mathematical approach provides a systematic way to tackle reflection problems in geometry.
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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?
 
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RTCNTC said:
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
 
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:

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

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

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

Hence:

$$mb'-mb=a-a'\tag{1}$$

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

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

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

$$ma+k-b=-ma'-k+b'$$

$$b+b'-k=ma+ma'+k\tag{2}$$

From (1), we find:

$$b'=\frac{mb+a-a'}{m}$$

And from (2):

$$b'=ma+ma'-b+2k$$

Hence:

$$\frac{mb+a-a'}{m}=ma+ma'-b+2k$$

Solving for $a'$, there results:

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

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

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

Now we have a general formula. (Clapping)

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

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

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

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

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

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

Hence:

$$mb'-mb=a-a'\tag{1}$$

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

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

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

$$ma+k-b=-ma'-k+b'$$

$$b+b'-k=ma+ma'+k\tag{2}$$

From (1), we find:

$$b'=\frac{mb+a-a'}{m}$$

And from (2):

$$b'=ma+ma'-b+2k$$

Hence:

$$\frac{mb+a-a'}{m}=ma+ma'-b+2k$$

Solving for $a'$, there results:

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

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

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

Now we have a general formula. (Clapping)

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

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

$$b'=\frac{(3^2-1)b+2((3)a+0)}{3^2+1}=\frac{8b+6a}{10}=\frac{4b+3a}{5}$$
Thank you very much. Great work!
 
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