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Anna35 [415]
3 years ago
13

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

sformed under the rule (x,y) (x-1,y-2) then does T' form a right angle at <GRT'?

Mathematics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

m=\frac{y_2-y_1}{x_2-x_1}

The product of slopes of two perpendicular lines is -1.

Slope of GR is

\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}

Slope of RT is

\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1

Product of slopes of GR and RT is

\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

(x,y)\rightarrow(x-1,y-2)

Then the coordinates of T' are

(2,6)\rightarrow(2-1,6-2)

(2,6)\rightarrow(1,4)

Slope of RT' is

\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}

Product of slopes of GR and RT' is

\frac{-4}{3}\times \frac{3}{4}=-1

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

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