Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5) and H(6, 5); where could points J and K be located? Group of answer choices J(6, 2); K(2, 2) J(6, 4); K(2, 4) J(6, 9); K(2, 9) J(6, 7); K(2, 7)

J (6,2) K (2,2)

Step-by-step explanation:

Now distance of AB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

                                = 8 units

CB =  [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 6 units by using same formula

Now AB/CB must equal to GH/HJ as rectangles are similar

GH =  [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 4 units

so

8/6 = 4/HJ

So,

HJ = 3 units

Now if we see the coordinates given carefully, it is obvious that two perpendicular lines lie perfectly parallel to x and y coordinates in rectangle ABCD.  A is (-7,-2) and B is (1,-2) which means distance along y-axis doesn't change. Similarly for C (1,-8) and D (-7,-8), one can see that distance between y-axis doesn't change. So lines AB and CD of rectangle are parallel with x and AD and BC are parallel with y-axis.

In rectangle GHJK one can see that in given coordinates, G(2,5) and H(6,5), y coordinate is same so it is parallel to x axis. Now, HJ is perpendicular to GH so it must be parallel to y axis. It means if we know the lengths of sides we can easily determine unknown coordinates by simple addition and subtraction.

So,  we know HJ = 3 units

J is (6,2) since HJ is parallel to y axis so distance on x axis will remain unchanged and length of line HJ will effect distance of y axis.

Similarly K is (2,2) for the same reason.