Question

Part A: Mrs. Konsdorf claims that angle R is a right angle. Is she correct? Explain your reasoning. Part B: If T is transformed under the rule (x,y) —> (x-1, y-2), then does T’ Form a right angle at angle GRT’?

Answer 1

Part A.

In order to see if R is a right angle, we need to find the slope of the line segments GR and RT.

The points G and R are

$\begin{gathered} G=(x_1,y_1)=(-6,5) \\ R=(x_2,y_2)=(-3,1) \end{gathered}$

By substituting these point into the slope formula ,we have

$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{1-5}{-3-(-6)} \end{gathered}$

which gives

$\begin{gathered} m=\frac{-4}{3} \\ m=-\frac{4}{3} \end{gathered}$

Now, lets find the slope of the line segment RT.

The points R and T are

$\begin{gathered} T=(x_1,y_1)=(2,6) \\ R=(x_2,y_2)=(-3,1) \end{gathered}$

then, the slope is

$\begin{gathered} M=\frac{y_2-y_1}{x_2-x_1} \\ M=\frac{1-6}{-3-2} \\ M=\frac{-5}{-5}=1 \end{gathered}$

Finally, two lines segments are perpendicular (or in other words, R is a right angle) if one of the slopes is a negative reciprocal of the other, that is, the following equality must be fulfilled:

$m=-\frac{1}{M}$

However, in our case, we can see that

$-\frac{4}{3}\ne-\frac{1}{1}$

Therefore, R is not a right angle.

Part B.

We want to see if angle R is a right angle but with a new T, which is now T'. In this case, we can apply the same procedure but with the point T' as

$T^{\prime}=(x_1,y_1)=(1,4)$

because T was (2,6) then T'=(2-1,6-2)=(1,4).

So, the new slope for the segment RT' is

$\begin{gathered} M=\frac{1-4}{-3-1} \\ M=\frac{-3}{-4}=\frac{3}{4} \end{gathered}$

But now, we can see that

$\begin{gathered} m=-\frac{1}{M} \\ \text{yields,} \\ -\frac{4}{3}=-\frac{1}{\frac{3}{4}} \\ -\frac{4}{3}=-\frac{4}{3} \end{gathered}$

since both number are the same, then m is the negative reciprocal of the new M, therefore, the line segments GR and RT' are perpendicular, which implies that R is a right angle.