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?
+114221 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" ) :
