find the length of the median of triangle whose vertices are 1 , - 1 , 0,2 and -5,3 ... plz can any one help ..........
Answers
We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,-1); B (2, 1) and C (0, 3)
So we should find the mid-points of the sides of the triangle.
In general to find the mid-point P(x,y) of two points
A
(
x
1
,
y
1
)
A(x1,y1) and
B
(
x
2
,
y
2
)
B(x2,y2) we use section formula as,
P
(
x
,
y
)
=
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
P(x,y)=(x1+x22,y1+y22)
Therefore mid-point P of side AB can be written as,
P
(
x
,
y
)
=
(
2
+
0
2
,
1
−
1
2
)
P(x,y)=(2+02,1-12)
Now equate the individual terms to get,
x = 1
y = 0
So co-ordinates of P is (1, 0)
Similarly mid-point Q of side BC can be written as
Q
(
x
,
y
)
=
(
2
+
0
2
,
3
+
1
2
)
Q(x,y)=(2+02,3+12)
Now equate the individual terms to get,
x = 1
y = 2
So co-ordinates of Q is (1, 2)
Similarly mid-point R of side AC can be written as,
R
(
x
,
y
)
=
(
0
+
0
2
,
3
−
1
2
)
R(x,y)=(0+02,3-12)
Now equate the individual terms to get,
x =1
y= 2
So co-ordinates of Q is (1, 2)
Similarly mid-point R of side AC can be written as,
R
(
x
,
y
)
=
(
0
+
0
2
,
3
−
1
2
)
R(x,y)=(0+02,3-12)
Now equate the individual terms to get,
x = 1
y = 1
So co-ordinates of R is (0, 1)
Therefore length of median from A to the side BC is,
A
Q
=
√
(
0
−
1
)
2
+
(
−
1
−
2
)
2
AQ=(0-1)2+(-1-2)2
=
√
1
+
9
=1+9
=
√
10
=10