What you have here is the reflection problem. All you have to do is reflect from the original points over the x-axis or the y-axis.
1.) A(2,2), B(5,4), C(5,1) You have to reflect over the x-axis. And over the axis is (+,-), so the y-intercept must be negative. And the answer to that is simple: A'(2,-2), B'(5,-4), C'(5,-1). For reflection, all you have to do change the signs depends what quarter you on: Q1(+,+), Q2(-,+), Q3(-,-), Q4(+,-).
2.) A(-1,2), B(2,-1), C(5,2), D(2,5) Now this is the part you need to pay close attention on. As you can see figure A is already over y-axis, but that doesn't mean you can't reflect it. It means you got the new y-axis, which is: (-2,0).
So the answer is this: A'(-3,2), B'(-4,-1), C'(-7,2), D'(-4,5).
3.) B(1,-2)=>B'(1,2) Remember what we discuss before, "...all you have to do is change the signs depends what quarter you on" And that actually what this showing us, now you have figure it out whether it reflect over x-axis or the y-axis. You notice that only the y-intercept had change the signs, so it must be: x-axis. Because only Q4 is reflecting to Q1.
4.) J(-3,5)=>J'(-3,-5) Okay, J is the point for Q2 and J' is for Q3. The answer is it reflect over the x-axis.
5.) W(-7,-4)=>W'(7,-4) W in Q3 and W' in Q4. It reflect over the y-axis.
6.) Same as we did for question 2, we got the new y-axis (which is (-1,0)). The question is asking us what is the when we reflect X(4,2) over y-axis. And the answer is X'(-5,2).
7.) Same as the previous question, Y(4,4) over y-axis and the answer is: Y'(-6,4).
8.) Z(0,2) the answer is Z'(-2,2).
9.) For this question, it asking us to graph the whole triangle that is reflect it from the original triangle over the x-axis. The correct answer is: X'(4,-2), Y'(4,-4), Z'(0,-2).
Well that's all folks!! LOL! But I hope this helps you alot.
