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# hard geometry graph transformations help

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Parallelogram $$ABCD$$ with A(2,5), B(4,9), C(6,5), and D(4,1) is reflected across the x-axis to $$A'B'C'D'$$ and then $$A'B'C'D'$$ is reflected across the line $$y = x + 1$$ to $$A''B''C''D''$$. This is done such that D' is the image of D, and D'' is the image of D'. What is the ordered pair of D'' in the coordinate plane?

Please help I don't know where to start. Jan 26, 2022
edited by proyaop  Jan 26, 2022
edited by proyaop  Jan 26, 2022

#1
+5

I got (-2,5) is that correct?

Jan 26, 2022
#2
+2

I graphed it just becasue that is my idea of fun.

But really you are only interested in D  the other points have no consequence.

It is sensible to do a rough graph of D so that you can see where you might go with it.

D(4,1)

reflect it across the x-axis and you get   D'(4,-1)

Now you want to reflect that across the line y=x+1

the gradient of that line is one the line at right angles will have a gradient of -1  so we want the line that passes thorough (4,-1) with a gradient of -1

y=-x+b

-1=-4+b

b=3

y=-x+3

I also want the perpendicular distance from  (4,-1) to   y=x+1

x-y+1=0

$$d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}\\ d=\frac{|4+1+1|}{\sqrt{1+1}}\\ d=\frac{6}{\sqrt2}\\ d=3\sqrt2$$

So I want the point on the line y=-x+3 that is 3sqrt2  from (4,-1)

$$(x-4)^2+(y+1)^2=72\quad \cap \quad y=-x+3\\ (x-4)^2+(-x+3+1)^2=72\\ (x-4)^2+(-x+4)^2=72\\ 2(x-4)^2=72\\ x-4=\pm\sqrt{36}\\ x=\pm6+4\\ x=-2 \;\;or \;\;10$$

I obviously want the negative one

x=-2,  y=--2+3 = 5

so D" is (-2,5)

Think about it first, but then if you have questions, make sure you ask.  Jan 26, 2022