Start by plotting the points of the vertices and connecting them.
See solution.
Practice makes perfect
We want to graph â–³ GHJ with vertices at G(0,1), H(4,0), and J(4,1). We will first plot the points of the vertices, then connect the points. Let's do it!
Now we can graph the image of â–³ GHJ after a translation of three units up followed by a reflection over the y-axis. We will start by translating â–³ GHJ three units up.
Next, we can reflect the image of translation, triangle G'H'J', over the y-axis.
Note that â–³ G''H''J'' is the image after the series of transformations series of transformations of a translation of three units up followed by a reflection over the y-axis. This means that we graphed the preimage and found the image after the series of transformations!
We can find the side lengths of both the preimage and the image. Notice that we can find the lengths of GJ, JH, G'J', and J'H' by using the grids in the graph.
We know that GH is a length. Since lengths cannot be negative, length of GH needs to be positive. We found the length of each side of â–³ GJH as follows.
GJ&=4 units
JH&=1 unit
GH&=sqrt(17) units
We know from the graph that J''G'' and J''H'' have the same lengths as JG and JH, respectively. If we use the Pythagorean Theorem in a similar way, we find that G''H'' has the same length as GH, sqrt(17) units.
G''J''&=4 units
J''H''&=1 unit
G''H''&=sqrt(17) units
Because the sizes and shapes of â–³GHJ and â–³G''H''J'' are the same, these triangles arecongruent.
