find the coordinates of the points of the trisection of the line segment joining (4,-1),(-2,-3)
Answers
We have two points with position vector and
and so the vector joining these two points is
We need to divide this vector by three, in order to find the points of trisections. So the vector becomes
The position vector of one point of trisection of the line segment is,
And that of the other point is,
Hence the points of trisections are and
Answer:
We have two points with position vector \left < 4,\ -1\right >⟨4, −1⟩ and \left < -2,\ -3\right > ,⟨−2, −3⟩, and so the vector joining these two points is \left < 4-(-2),\ -1-(-3)\right > =\left < 6,\ 2\right > .⟨4−(−2), −1−(−3)⟩=⟨6, 2⟩.
We need to divide this vector by three, in order to find the points of trisections. So the vector becomes \dfrac{1}{3}\left < 6,\ 2\right > =\left < 2,\ \dfrac{2}{3}\right > .
3
1
⟨6, 2⟩=⟨2,
3
2
⟩.
The position vector of one point of trisection of the line segment is,
\longrightarrow\vec{t_1}=\left < -2,\ -3\right > +\left < 2,\ \dfrac{2}{3}\right >⟶
t
1
=⟨−2, −3⟩+⟨2,
3
2
⟩
\longrightarrow\vec{t_1}=\left < 0,\ -\dfrac{7}{3}\right >⟶
t
1
=⟨0, −
3
7
⟩
And that of the other point is,
\longrightarrow\vec{t_2}=\left < 0,\ -\dfrac{7}{3}\right > +\left < 2,\ \dfrac{2}{3}\right >⟶
t
2
=⟨0, −
3
7
⟩+⟨2,
3
2
⟩
\longrightarrow\vec{t_2}=\left < 2,\ -\dfrac{5}{3}\right >⟶
t
2
=⟨2, −
3
5
⟩
Hence the points of trisections are \bf{\left(0,\ -\dfrac{7}{3}\right)}(0, −
3
7
) and \bf{\left(2,\ -\dfrac{5}{3}\right)}.(2, −
3
5
).
Step-by-step explanation:
Hope this answer will help you.