Answer to Question #90622 in Algebra for Laura

Question

Rectangle PQRS has vertices P(1,4) Q(6,4) R(6,1) and S(1,1) without graphing find the new cooedinates B and C of ABC are (2,6) and (4,3) and coordinates B and C of its image are (-2,2) and (1,4) what coordinates of vertex A

Answer 1

Distances from (2,6) to (4,3) and from (-2,2) to (1,4) are the same: sqrt((2-4)^2+(6-3)^2)=sqrt(4+9)=sqrt(13), sqrt((-2-1)^2+(4-2)^2)=sqrt(9+4)=sqrt(13). So we can state that transformation does not change length of the sides of ABC, but only changes its location. So it’s a linear transform (new x is equals to kx+b, new y is equals to my+n). Let’s find k, b, m, n by putting B and C values into the equations:

2k+b=-2, 4k+b=1, 2k=3, k = 3/2.

b = -2 – 2k = -2-3 = -5.

6m + n = 2, 3m+n=4, 3m=-2, m=-2/3.

n = 2-6m = 2-6*(-2/3)=2+4=6.

So transformation looks like (3/2*x-5, 2/3*y+6). Substituting values of coordinates of the point A we can compute coordinates of the image of point A. For example, let A be (1,1). In this case the image of A will be (3/2*1-5,2/3*1+6) = (-7/2,20/3).