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Parallelogram ABCD with (2,5), (4,9), (6,5), and (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?

CluelesssPersonnn  Jul 11, 2018
 #1
avatar+91019 
+3

The original D  is  ( 4,1)

 

When reflected across the x axis, 'D'   is  ( 4, -1)

 

D' and  D" will lie on a line that  will be  perpendicular to  y  = x + 1

So....the slope of this line is  - 1

And the equation of this line   is

y  = - ( x - 4) - 1

y = -x + 4 - 1

y = -x + 3

 

The  intersection  of these two lines  will be the midpoint  of   D'  and D"

So...setting the y's equal, we have

-x + 3  = x + 1       add  x to both sides,  subtract 1 from each side

2   = 2x     divide both sides by 2

1  = x

 

And using either line to  find the y coordinate of this intersection point, we have that

y = x + 1

y = 1 + 1

y  = 2

So...the midpoint  of  D' and D"  is  ( 1, 2 )

 

So...using the midpoint formula with the points ( 4, -1) and (1,2), we can  find  D'  as

(x + 4)/2  =   1                                ( y + -1) / 2  = 2

Multiply  both sides  by 2

x + 4  = 2                                           y - 1  = 4

subtract 4 from both sides                 add 1 to each side

 

x = -2                                                 y = 5

 

So...  D"    =  (-2, 5)

 

Here's a graph that shows this :

 

https://www.desmos.com/calculator/rbi6tjxihp

 

 

 

cool cool cool

CPhill  Jul 11, 2018
edited by CPhill  Jul 11, 2018
edited by CPhill  Jul 11, 2018
 #2
avatar+25 
0

THX!!!!!!!

CluelesssPersonnn  Jul 12, 2018

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