+0  
 
0
57
1
avatar

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+5 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?

 Dec 7, 2020
 #1
avatar+114221 
+1

This problem is lengthy......I will show you how to  get A".....the  calculation of the other three points  follow a similar procedure....incidentally......this problem is  better  done with a app  like Geogebra  (it unburdens the calculations)

 

A = (2,5)       A'  =  (2,-5)

 

Equation of a line  perpendicular  to   y = x +  5  is given by

 

y =  -(x -2)  - 5

 

Find the  x intersection of these two lines

 

-(x - 2)  - 5 =   x + 5

-x -3  = x + 5

-2x = 8

x = -4

And  y = (-4) + 5 =  1

 

So....the intersection of  these two lines =   (-4,1)

Distance^2  from  A'  to  this point   =  (-4-2)^2  + ( -5, -1)^2  =  72

 

So....constructing  a  circle  with a radius^2 of  72  centered  at  (-4,1)   gives us the equation

(x + 4)^2  + ( y -1)^2  = 72

Finding the intersection  of  this circle with the  line  y = -(x - 2) - 5    gives us A"  = ( -10,7)

 

So....if you follow this through you  should find that

 

A" = ( -10,7)     B"  = (-14,9)   C"  = (-10,11)   D"  = (-6,9)

 

Here's  an image  (the point (-4,1) and  circle shown is that  used to  get A" )  :

 

 

 

 

 

 

 

cool cool cool

 Dec 7, 2020
edited by CPhill  Dec 7, 2020

27 Online Users

avatar
avatar
avatar