Question

The vertices of triangle ABC are given: A (0;-1) B(12; -10) C(10;4)

Find:

a) the equation of the median and bisector drawn from vertex A;

b) the equation of the altitude from vertex B;

c) the point of intersection of the median and height;

d) the coordinates of the center of mass (the point of intersection of the medians).

Answer 1

Solution
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Median bisects the line.
D≡(
2
−3+5
​
,
2
−9−8
​
)≡(1,
2
−17
​
)
E≡(−
2
1+5
​
,
2
6−8
​
)≡(2,−1)
F≡(−
2
1−3
​
,
2
6−9
​
)≡(−2,
2
−3
​
)
∴ equation of AD:(y−y
1
​
)=(
x
2
​
−x
1
​

y
2
​
−y
1
​

​
)
AD
​
(x−x
1
​
)
⇒(y−6)=
⎝
⎜
⎜
⎛
​

1+1
2
−17
​
−6
​

⎠
⎟
⎟
⎞
​
(x+1)⇒(y−6)=−
4
29
​
(x+1)
⇒
4
29
​
x+y−6+
4
29
​
=0
AD⇒29x+4y+5=0
equation of BE:(y+9)=(−
2+3
1+9
​
)
BE
​
(x+3)
⇒(y+9)=
5
8
​
(x+3)⇒5y+45=8x+24
BE⇒8x−5y−21=0
Equation of CF:(y+8)=
⎝
⎜
⎜
⎛
​

−2−5
−
2
3
​
+8
​

⎠
⎟
⎟
⎞
​

CF
​
(x−5)
⇒(y+8)=−
14
13
​
(x−5)⇒14y+112=−13x+65
CF:13x+14y+47=0