Question

URGENT! PLS HELP!
Find the lengths of the medians of a Triangle ABC whose vertices are A(8, –8), B(6, 8) C(2, 4).

Answer 1

Answer:

$m_{a}$ = 2√53 ≈ 14.56 ; $m_{b}$ = √101 ≈ 10.05 ; $m_{c}$ = √41 ≈ 6.40 ;

Step-by-step explanation:

A(8, –8), B(6, 8), C(2, 4)

$midpoint_{AB} ( \frac{x_{A}+x_{B} }{2}, \frac{y_{A} +y_{B}}{2})$ = $(\frac{8+6}{2} ,\frac{-8+8}{2})$ = $M_{C}$ (7, 0) ;

midpoint of BC = [(6 + 2)/2, (8 + 4)/2] = $M_{A}$ (4, 6) ;

midpoint of AC = [(8 + 2)/2, (- 8 + 4)/2] = $M_{B}$ (5, - 2) ;

$d=\sqrt{(x_{2} -x_{1})^2 + (y_{2} -y_{1})^2}$

$CM_{C}$ = $\sqrt{(7-2)^2 +(0-4)^2}$ = √41 ≈ 6.40

$AM_{A}$ = √[(4 - 8)² + (6 + 8)²] = 2√53 ≈ 14.56

$BM_{B}$ = √[(5 - 6)² + (- 2 - 8)²] = √101 ≈ 10.05

$m_{a}$ = 2√53 ≈ 14.56 ; $m_{b}$ = √101 ≈ 10.05 ; $m_{c}$ = √41 ≈ 6.40 ;