Question

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

Answer 1

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:
$d=\sqrt{(3.5-6.5)^2+(5.5-5.5)^2} \\=\sqrt{(-3)^2+(0)^2} \\=\sqrt{9+0}=\sqrt{9}=3$