Find the length of the median of triangle abc whose vertices are a(0,-1) b (2,1) and c(0,3)
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In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²
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Step-by-step explanation:
Given:
\text{In $\triangle$ABC, A(0,-1), B(2,1) and C(0,3)}In △ABC, A(0,-1), B(2,1) and C(0,3)
\textbf{To find:}To find:
\text{Lengths of the medians of $\triangle$ABC}Lengths of the medians of △ABC
\textbf{Solution:}Solution:
\text{Let D, E and F be the midpoints of sides BC, AC and AB}Let D, E and F be the midpoints of sides BC, AC and AB
\text{By Using Midpoint formula,}By Using Midpoint formula,
\text{Midpoint of BC}Midpoint of BC
=(\dfrac{2+0}{2},\dfrac{1+3}{2})=(
2
2+0
,
2
1+3
)
=\bf\,D(1,2)=D(1,2)
\text{Midpoint of AC}Midpoint of AC
=(\dfrac{0+0}{2},\dfrac{-1+3}{2})=(
2
0+0
,
2
−1+3
)
=\bf\,E(0,1)=E(0,1)
\text{Midpoint of AB}Midpoint of AB
=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})=(
2
x
1
+x
2
,
2
y
1
+y
2
)
=(\dfrac{0+2}{2},\dfrac{-1+1}{2})=(
2
0+2
,
2
−1+1
)
=\bf\,F(1,0)=F(1,0)
\textbf{Length of the median AD}Length of the median AD
=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
=\sqrt{(0-1)^2+(-1-2)^2}=
(0−1)
2
+(−1−2)
2
=\sqrt{1+9}=
1+9
=\sqrt{10}=
10
\textbf{Length of the median BE}Length of the median BE
=\sqrt{(2-0)^2+(1-1)^2}=
(2−0)
2
+(1−1)
2
=\sqrt{4+0}=
4+0
=\sqrt{4}=
4
=2=2
\textbf{Length of the median CF}Length of the median CF
=\sqrt{(0-1)^2+(3-0)^2}=
(0−1)
2
+(3−0)
2
=\sqrt{1+9}=
1+9
=\sqrt{10}=
10
\therefore\textbf{The length of the medians AD, BE and CF are $\bf\sqrt{10}$, 2 and $\bf\sqrt{10}$}∴The length of the medians AD, BE and CF are
10
, 2 and
10
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In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²