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=(y_2-y_1)/(x_2-x_1) \\ m=(1-5)/(-3-(-6)) \end{gathered}`

which gives

`\begin{gathered} m=(-4)/(3) \\ m=-(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=(y_2-y_1)/(x_2-x_1) \\ M=(1-6)/(-3-2) \\ M=(-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=-(1)/(M)`

However, in our case, we can see that

`-(4)/(3)\\e-(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=(1-4)/(-3-1) \\ M=(-3)/(-4)=(3)/(4) \end{gathered}`

But now, we can see that

`\begin{gathered} m=-(1)/(M) \\ \text{yields,} \\ -(4)/(3)=-(1)/((3)/(4)) \\ -(4)/(3)=-(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.