Abc 3 r

Asked by maham237 @ 08/01/2021 in Mathematics viewed by 324 People

Given △ABC where A(2, 3), B(5, 8), C(8, 3), RS is the mid segment parallel to AC, ST is the mid
segment parallel to AB, and RT is the mid segment parallel to BC, determine if the statements are true or false.Select True or False for each statement

Answered by maham237 @ 08/01/2021

Since RS is a midsegment parallel to AC, that means R is the midpoint of AB and S is the midpoint of BC. The midpoint formula is:
$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ . Using the coordinates of A and B, we have:
$R=(\frac{5+2}{2},\frac{8+3}{2})
\\R=(\frac{7}{2},\frac{11}{2})
\\R=(3.5, 5.5)$ . Similarly, S is the midpoint of BC:
$S=(\frac{5+8}{2},\frac{8+3}{2})
\\S=(\frac{13}{2},\frac{11}{2})
\\S=(6.5, 5.5)$ .
Since ST is a midsegment parallel to AB, then T must be a midpoint of AC:
$T=(\frac{2+8}{2},\frac{3+3}{2})
\\T=(\frac{10}{2},\frac{6}{2})
\\T=(5,3)$ .
Now that we have the coordinates of each point we can find the length of each segment using the distance formula:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
For ST:
$d=\sqrt{(6.5-5)^2+(5.5-3)^2}
\\=\sqrt{(1.5)^2+(2.5)^2}
\\=\sqrt{2.25+6.25

\\=\sqrt{8.5}=2.9 \neq 4$
For RT:
$d=\sqrt{(3.5-5)^2+(5.5-3)^2}
\\=\sqrt{(-1.5)^2+(2.5)^2}
\\=\sqrt{2.25+6.25}
\\=\sqrt{8.5}=2.9 \neq 5$
For RS:
$d=\sqrt{(3.5-6.5)^2+(5.5-5.5)^2}
\\=\sqrt{(-3)^2+(0)^2}
\\=\sqrt{9+0}=\sqrt{9}=3$

Abc 3 r

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