Question

Triangle ABC is translated on the coordinate plane below to create triangle A'B'C':

Triangle ABC, triangle A prime B prime C prime, and parallelogram EFGH on the coordinate plane with ordered pairs at A negative 7, 6, at B negative 5, 3, at C negative 1, 3, at A prime 1, 4, at B prime 3, 1, at C prime 7, 1, at E negative 5, negative 2, at F negative 1, negative 2, at G negative 3, negative 5, at H negative 7, negative 5

If parallelogram EFGH is translated according to the same rule that translated triangle ABC, what is the ordered pair of point H'?

(−15, −3)

(−9, 3)

(1, −7)

(−5, −13)

Answer 1

First you need to find the rule or 'equation' that this system goes by before we can solve what H' point is located.

So, to do so, label out the triangles points:

A(-7, 6), B(-5, 3), C(-1, 3) to A'(1, 4), B'(3, 1), C'(7, 1)

Now for A to A' which is -7 to 1 is a 8 point difference and A to A' which is 6 to 4 is a -2 point difference, indicating:
(x + 8, y - 2)

Now:

B(-5, 3) -> (-5 + 8, 3 - 2) = B'(3, 1)
C(-1, 3) -> (-1 + 8, 3 - 2) = C'(7, 1)

Now to simply solve for H to H':

We know H 's coordinates are: (-7, -5)

Now add in the rule and solve:

(-7 + 8, -5 - 2) = ?

Answer: (1, -7)