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Oxana [17]
3 years ago
10

For ΔABC with A(–5, –4), B(3, –2), and C(–1, 6), M is the midpoint of

Mathematics
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

The proposition is true, as \overrightarrow{MN} = (2,5) and \overrightarrow {AC} = (4, 10) satisfies the given expression (\overrightarrow{MN} = \frac{1}{2}\cdot \overrightarrow{AC}).

Step-by-step explanation:

Vertices of the triangle ABC are located in A(x,y) = (-5,-4), B(x,y) = (3,-2) and C(x,y) = (-1, 6), respectively. By definition of midpoint, we get that M and N are located in the following coordinates:

M(x,y) = \frac{1}{2}\cdot A(x,y) +\frac{1}{2}\cdot B(x,y) (Eq. 1)

M(x,y) = \frac{1}{2}\cdot (-5,-4)+\frac{1}{2}\cdot (3,-2)

M(x,y) = \left(-\frac{5}{2},-2 \right)+\left(\frac{3}{2},-1\right)

M(x,y) = (-1,-3)

N(x,y) = \frac{1}{2}\cdot B(x,y) + \frac{1}{2}\cdot C(x,y) (Eq. 2)

N(x,y) = \frac{1}{2}\cdot (3,-2)+\frac{1}{2}\cdot (-1,6)

N(x,y) = \left(\frac{3}{2},-1\right)+\left(-\frac{1}{2},3 \right)

N(x,y) = (1, 2)

Then, the relative vectors between A and C, as well as M and N are, respectively:

\overrightarrow{AC} = C(x,y)-A(x,y) (Eq. 3)

\overrightarrow{AC} = (-1, 6) -(-5,-4)

\overrightarrow {AC} = (4, 10)

\overrightarrow {MN} = N(x,y)-M(x,y) (Eq. 4)

\overrightarrow {MN} = (1,2)-(-1,-3)

\overrightarrow{MN} = (2,5)

By getting back into (Eq. 3), we find the following the result after some algebraic handling:

\overrightarrow {AC} = (4, 10)

\overrightarrow{AC} = 2\cdot (2,5)

\overrightarrow {AC} = 2\cdot \overrightarrow {MN}

\overrightarrow{MN} = \frac{1}{2}\cdot \overrightarrow{AC}

Which proofs the given affirmation.

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