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lets say for example we want to reflect point (-6,-4) which is A 

and lets say we want to reflect it across like line y=-4x-4 


how would you find A'??

thank you so much i dont understand how this works :_((

 Oct 18, 2019
 #1
avatar+105664 
+3

lets say for example we want to reflect point (-6,-4) which is A 
and lets say we want to reflect it across like line y=-4x-4 

 

Well you could probably do it graphically.

or

 

 

You could say that the gradient of the line is -4 so the gradient of the perpendicular is 1/4

The equation of the perpendicular through (-6,-4) is    

y=0.25x+k

-4=0.25*-6+k

-4=-1.5+k

k=-2.5

so

y=0.25x-2.5

 

simulaneous solution of the two lines

\(y=-4x-4 \\ y=0.25x-2.5\\ -4x-4=0.25x-2.5\\ -1.5=4.25x\\ x=-6/17\\ y=-2\frac{10}{17}\\ (\frac{-6}{17},-2\frac{10}{17})\)

 

 

Now you can find the distance between those two points, then find the other point on the perpendicular that is on other side of the reflection line and the same distance away as the original point was.

 

These numbers are horrible.  Did you make this question up?

 Oct 18, 2019
 #3
avatar+656 
+1

well not really, i asked my friend and she gave me this example...
blush hehe i know they're really ugly numbers 


but when i did it i got like -0.4 for the y value and 5.2 for the x value for the midpoint, but i probably made careless calculation errors... 

anywho... 
thank you melody so much for your help <3 like actually i forgot how to do these completely :')

 

Nirvana  Oct 19, 2019
 #5
avatar+105664 
+1

It could have been me that made the careless error.  I did not check my answer.

I often leave it to other people to pick up my silly mistakes.

Melody  Oct 19, 2019
 #6
avatar+656 
+1

but let me ask you this...

what is a gradient in mathematics?
sorry lol im dumb :')

Nirvana  Oct 19, 2019
 #7
avatar+105664 
+2

Hi Nirvana,

Asking genuine questions is never dumb.      BUT     not asking them is always dumb.

 

____________________

 

The gradient is the slope of a line.

 

Have a look through the explanation on this site.

 

https://www.mathsisfun.com/gradient.html

Melody  Oct 19, 2019
edited by Melody  Oct 20, 2019
 #8
avatar+105664 
+2

Here is the graph that we were talking about

 

Melody  Oct 20, 2019
 #9
avatar+656 
+1

rip :') 
well thank you so much melody!! <33
i never knew that we could call the slope of a line a gradient 

thank you for clarifying 

Nirvana  Oct 20, 2019
 #2
avatar+104911 
+2

 

To continue  ......

 

We know  that   the intersection of the two lines  is  ( -6/17,  -2  10 /17)    = (-6/17,  -44/17)

 

This is the midpoint between  A  and A'

 

So......to find A' = (x, y)  we have that

 

 [-6 + x ] / 2  = -6/17              [ -4 + y ] / 2   =  -44/17 

-6 + x  =   -12/17                     -4 +  y   =   -88/17

x  =  -12/17 + 6                         y  =  -88/17 + 4

x =  -12/17                                y =  -20/17

x  = 90/17

 

So  A'  =  (90/17, -20/17)

 

Here's a graph  :  https://www.desmos.com/calculator/trsmplhqlg   

 

 

cool cool cool

 Oct 18, 2019
 #4
avatar+656 
+1

thank you so much CPhill! <3

Nirvana  Oct 19, 2019
 #10
avatar+656 
+1

@Melody and @CPhill 

Let me ask you this though,,, (i'm sorry if i keep asking questions btw) 
Is there a direct formula for finding the new point that's been reflected off another point? 

smiley thank you!

 Oct 20, 2019
 #11
avatar+105664 
+2

You should not appologize for asking good questions on a help site.  wink

 

I could work one out but there is none that are normally taught.

A mathematician can always work stuff like that out. 

The formula would be needed if you were designing a computer program to sketch reflections such as I used for the diagram above.

 

It would be a bit complicated though.

 

If the points and line are easier then you can do it just by visual inspection.

Melody  Oct 21, 2019
 #12
avatar+656 
+1

ah, i see :) 
thanks

Nirvana  Oct 21, 2019

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