Learn how to find the image of a given reflection.
Log in Hannah 6 years agoPosted 6 years ago. Direct link to Hannah's post “I understand how to algeb...” I understand how to algebraically perform reflections if the line of reflection is y = 0, x = 0, y = x, or y = -x. How can I algebraically perform a reflection for ANY line of reflection (e.g. how could I reflect (2, 9) across y = 7x + 2 algebraically)? • (39 votes) Ian Pulizzotto 6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “Great question!Let A be...” Great question! Let A be the point to be reflected, let k be the line about which the point is reflected, let B represent the desired point (image), and let C represent the intersection of line k and line AB. Note that line AB must be perpendicular to line k, and C must be the midpoint of segment AB (from the definition of a reflection). Example: suppose we want to reflect the point A(2,9) about the line k with equation y = 7x + 2. So we first find the equation of the line through (2,9) that is perpendicular to the line y = 7x + 2. Since the line y = 7x + 2 has slope 7, the desired line (that is, line AB) has slope -1/7 as well as passing through (2,9). So the desired line has an equation of the form y = (-1/7)x + b. Substituting the point (2,9) gives Now we need to find the intersection of the lines y = 7x + 2 and y = (-1/7)x + 65/7 by solving this system of equations. So the intersection of the two lines is the point C(51/50, 457/50). Recall that A is the point (2,9). So the image (that is, point B) is the point (1/25, 232/25). (99 votes) natalie.stringer22 6 years agoPosted 6 years ago. Direct link to natalie.stringer22's post “I do not understand any o...” I do not understand any of this at all. Is there an easier way to learn/understand it? • (43 votes) 1658628926 a year agoPosted a year ago. Direct link to 1658628926's post “Simple reflections are a ...” Simple reflections are a matter of looking at a line and a point, line, or polygon on one side of it. You want to figure out the distance between the two and take that point, line, or polygon and put it the distance away from the line, but on the opposite side. Also, the line from any point A to its image A' is perpendicular to the line of reflection. Hope this helps! (11 votes) louisaandgreta 4 years agoPosted 4 years ago. Direct link to louisaandgreta's post “I could really use Sal ma...” I could really use Sal making a video about this, what’s written on this doc is really confusing. • (28 votes) yashaa 2 years agoPosted 2 years ago. Direct link to yashaa's post “The reflection line is th...” The reflection line is the line that you are reflecting over. Y=mx+b is just the basic slope-intercept equation. If you don't understand slope -intercept, I recommend watching the videos Khan provides in the algebra courses. Since geometry tends to be taught after algebra in some cases, I think it's why they didn't explain it more in depth. Hope this helps! (9 votes) McLachlin, Abigail a year agoPosted a year ago. Direct link to McLachlin, Abigail's post “I cried over this lesson ...” I cried over this lesson for over an hour, took a 2 day break,and then cried at the thought of it. I finally went back to it with a fresh head and realized I had over thought the whole things and it really wasn't that deep lol • (26 votes) 28jkim a year agoPosted a year ago. Direct link to 28jkim's post “wow that is bonkers” wow that is bonkers (5 votes) rl0262 6 years agoPosted 6 years ago. Direct link to rl0262's post “Is there a formula for th...” Is there a formula for the reflections? • (22 votes) 💲⚔💎💝Max Lennon💝💎⚔💲 2 years agoPosted 2 years ago. Direct link to 💲⚔💎💝Max Lennon💝💎⚔💲's post “count the spaces between ...” count the spaces between the line you are reflecting over (4 votes) andrewcwitt a year agoPosted a year ago. Direct link to andrewcwitt's post “Is there a more mathemati...” Is there a more mathematical way of calculating the reflection as opposed to manually counting on a graph? Perhaps using point slope (y=mx+b) or maybe by setting up a function? It seems like there should be a way to do this without requiring the graph. • (8 votes) connor N. a year agoPosted a year ago. Direct link to connor N.'s post “Yep, just plug the coordi...” Yep, just plug the coordinates for each point into the point-slope equation that is given to get the reflected points. Eg. In the last example in the article, you have the points M, O and N. To use M as an example, it's graphed at (-7,2). In the question, it tells you that it is reflected over a line of the equation y=-1-x To find the reflected coordinates of M using the point-slope equation given, plug in the following and solve: 2=-1-x (Gives you the X coordinate) You can repeat this step for the other points (O and N) and then graph it. Hope that helps. (17 votes) skylar.schrage 4 years agoPosted 4 years ago. Direct link to skylar.schrage's post “why cant there be a video...” why cant there be a video on this i dont understand it but a video would help • (14 votes) ananyanair378 4 years agoPosted 4 years ago. Direct link to ananyanair378's post “There is a part that says...” There is a part that says "I want to see Sal doing a similar question" which helped me since I was having trouble. (2 votes) Anthony Senner 2 years agoPosted 2 years ago. Direct link to Anthony Senner's post “Seriously, this math stum...” Seriously, this math stumps me • (12 votes) aoya joha 3 years agoPosted 3 years ago. Direct link to aoya joha's post “isn't there an algebraic ...” isn't there an algebraic formula for this ? • (4 votes) Yagnesh Peddatimmareddy 3 years agoPosted 3 years ago. Direct link to Yagnesh Peddatimmareddy's post “When you reflect a point ...” When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places. If you reflect over the line y = -x, the x-coordinate and y-coordinate change places and are negated (the signs are changed). the line y = x is the point (y, x). the line y = -x is the point (-y, -x). (11 votes) UmaFaLeung 5 years agoPosted 5 years ago. Direct link to UmaFaLeung's post “This is a probably a stup...” This is a probably a stupid question, but i totally do not get why -- for example -- in this problem: Draw the image of triangle MNO under a reflection over: y = -1 -x I don't get what y = -1 -x means, is it a coordinate or is it comparing y to x, i just don't understand, same goes for the other problems: What is the image of (-12, 12) under a reflection over line y = x. I can usually solve the problem, but i feel like i still need to understand what is means. • (5 votes) Sarah 5 years agoPosted 5 years ago. Direct link to Sarah's post “No questions are stupid! ...” No questions are stupid! y=-1-x and y=x are both lines. When you reflect a point, it is an equal distance away from the line as your original point. For instance, (-12,12) reflected over y=x would be (12,-12). I hope this clears things up! (9 votes)Want to join the conversation?
So we can first find the equation of the line through point A that is perpendicular to line k. Then we can algebraically find point C, which is the intersection of these two lines. Then, using the fact that C is the midpoint of segment AB, we can finally determine point B.
9 = (-1/7)(2) + b which gives b = 65/7. So the equation of this line is y = (-1/7)x + 65/7.
Using the substitution method gives 7x + 2 = (-1/7)x + 65/7; (50/7)x = 51/7; x = 51/50.
Then y = 7(51/50) + 2 = 457/50.
Since C is the midpoint of AB, we have
B = C + (C - A) = (51/50 + 51/50 - 2, 457/50 + 457/50 - 9) = (1/25, 232/25).
Sometimes they explain things that are pretty basic and other times more complicated things they’ll just assume that we know them even though we haven’t covered it/them yet.
For instance I don’t understand what they mean when referring to the reflection line
Y=1-x
Y=x+2
Y= x-5
y=-1-(-7) (Gives you the Y coordinate)
