<ol>
<li><p>P to P' (3,-1) by a glide reflection for the translation (x,y)
to (x-3, y) and reflection across the line y=2. What are the coordinates of P?
(HINT: Work backwards to find the pre-image coordinates)</p></li>
<li><p>What is the reflection image of (a,b) across the line y=-6?
A. (a-6, b) B. (a, b-6) C. (-12-a, b) D. (a, -12-b)</p></li>
</ol>
<p>Can you explain this to me? I know how to do it but I'm not sure how to explain it to my friend I am helping. Can somebody break it down with detail? Thanks in advance!</p>
<ol>
<li><p>I don’t understand the nomenclature.</p></li>
<li><p>You’re flipping a point (a, b) across a line parallel to the x-axis. If you draw a picture, you will find that coordinate a will stay the same for both the flipped point and the point itself. What will b be? Imagine the point (3,4). Now just consider the y coordinates y = 4 and y = -6. y = -6 is the midpoint of the segment formed by (a, b) and (a’, b’), where ’ denotes the flipped point. In this case, when b = 4, b’ must be -16.</p></li>
</ol>
<p>Since our previous condition eliminates answer choices A and C, we can just test the y portion of answer choices B and D.
First, B: (a, b-6) = (3,-2) - incorrect.
Now, D: (a, -12-b) = (3,-16) - ah, that’s what we determined the answer would be!</p>
<p>Brilliant! The answer is D (a, -12-b)!</p>
<p>For #2, if I’m not mistaken, it’d D. because all the other choices just don’t work anyway (and I do believe D. works, anyway, as well).</p>
<p>If you have to reflect an image across a line that has the equation y=-6, you have to first imagine any line or shape (that’s obviously made up of lines) on some positive (or negative, above -6) Y of the graph. Instead of working directly with numbers and eliminating that way to see if some line or shape ends up where it should for a reflection across the line y=-6, you can simply realize that if the line or shape you imagined was wayyyy in narnia somewhere, like with the coordinate of 50y, it must definitely not be reflectable upon y=-6 by just taking away 6 from it, as B. does. And, most obviously, taking away from the X coordinate won’t do us any good because we’re trying to reflect across a line going straight in the left/right direction, a.k.a. a line with slope 0, basically having us eliminate A. and C.</p>
<p>So, D. is the only one left. Besides, if you really wanted to, you could plug in some simple numbers like b=0.</p>
<p>tl;dr: You’re reflecting across a line of 0 slope, so there shouldn’t be any difference in X. A. and C. out. If a line of 0 slope as well had y=0, you can’t simply -6 from it to get the right reflection. Just D. left</p>
<p>If I understand what you are saying in #1…You are looking for the starting point so that after you apply the translation (x-3,y) – which translates three units left – and then take that resulting point and reflect it around the line y = 2, you end up at 3, -1. So to figure out the starting point, as the hint said, undo the two processes: Reflect back from 3, -1 around the y=2 line and you end up at 3, 5 – draw it and you will see that (3,-1) was three units below the reflection line, so now you want to be 3 units above. </p>
<p>But now, from (3,5) you have to translate three units to the right to undo the first translation. So that takes you to (6,5) where it all must have started…</p>
<p>That’s if I read this right. And of course…not an SAT question.</p>
<p>Thanks everyone! Yeah I thought she was asking about the SAT but it was just normal geometry.</p>