Question

45 (Two yertices of right triangle PQR are shown on the coordinate plane below. G) (t, fel 5 p=2,-6 2 3 4 5 6 7 823) Р 3 2 Q=2,3 6-5 -4 -3 -2 2 3 4 5 6 Q3 WN 8x2016 (- -) -2 -3 4 5 6| C-st) Qw ( Part A: What is the length, in units, of side PQ? Answer: 169 units Vertex R is located at (3,-2). Part B: What is the area, in square units, of triangle PQR? Show or explain how you know.

Answer 1

Given vertices have the next coordinates (x,y):

P(-6,2)

Q(3,2)

PART A: As the given vertices have the same y-coordinate, the distance is the difference between x-coordinates:

$\begin{gathered} PQ=3-(-6) \\ PQ=3+6 \\ PQ=9 \end{gathered}$

Then, the length of side PQ is 9 units

PART B: To find the area of the trinagle use the vertex R and vertex Q (they share the same x-coordinate) to find the height of the triangle:

The length of side QR is equal to the difference between y-coodinates of the vertices Q and R:

$\begin{gathered} QR=2-(-2) \\ QR=2+2 \\ QR=4 \end{gathered}$

As you can see in the calculations above and in the image the length of he base of the triangle (side PQ) is 9 units, and its hight (Side QR) is 4 units; use the base and hight to find the area as follow:

$\begin{gathered} A=\frac{1}{2}b*h \\ \\ A=\frac{1}{2}(9u)(4u) \\ \\ A=\frac{1}{2}(36u^2) \\ \\ A=18u^2 \end{gathered}$Then, the area of the given triangle is 18 square units