Answer:
The proposition is true, as [tex]\overrightarrow{MN} = (2,5)[/tex] and [tex]\overrightarrow {AC} = (4, 10)[/tex] satisfies the given expression ([tex]\overrightarrow{MN} = \frac{1}{2}\cdot \overrightarrow{AC}[/tex]).
Step-by-step explanation:
Vertices of the triangle ABC are located in [tex]A(x,y) = (-5,-4)[/tex], [tex]B(x,y) = (3,-2)[/tex] and [tex]C(x,y) = (-1, 6)[/tex], respectively. By definition of midpoint, we get that M and N are located in the following coordinates:
[tex]M(x,y) = \frac{1}{2}\cdot A(x,y) +\frac{1}{2}\cdot B(x,y)[/tex] (Eq. 1)
[tex]M(x,y) = \frac{1}{2}\cdot (-5,-4)+\frac{1}{2}\cdot (3,-2)[/tex]
[tex]M(x,y) = \left(-\frac{5}{2},-2 \right)+\left(\frac{3}{2},-1\right)[/tex]
[tex]M(x,y) = (-1,-3)[/tex]
[tex]N(x,y) = \frac{1}{2}\cdot B(x,y) + \frac{1}{2}\cdot C(x,y)[/tex] (Eq. 2)
[tex]N(x,y) = \frac{1}{2}\cdot (3,-2)+\frac{1}{2}\cdot (-1,6)[/tex]
[tex]N(x,y) = \left(\frac{3}{2},-1\right)+\left(-\frac{1}{2},3 \right)[/tex]
[tex]N(x,y) = (1, 2)[/tex]
Then, the relative vectors between A and C, as well as M and N are, respectively:
[tex]\overrightarrow{AC} = C(x,y)-A(x,y)[/tex] (Eq. 3)
[tex]\overrightarrow{AC} = (-1, 6) -(-5,-4)[/tex]
[tex]\overrightarrow {AC} = (4, 10)[/tex]
[tex]\overrightarrow {MN} = N(x,y)-M(x,y)[/tex] (Eq. 4)
[tex]\overrightarrow {MN} = (1,2)-(-1,-3)[/tex]
[tex]\overrightarrow{MN} = (2,5)[/tex]
By getting back into (Eq. 3), we find the following the result after some algebraic handling:
[tex]\overrightarrow {AC} = (4, 10)[/tex]
[tex]\overrightarrow{AC} = 2\cdot (2,5)[/tex]
[tex]\overrightarrow {AC} = 2\cdot \overrightarrow {MN}[/tex]
[tex]\overrightarrow{MN} = \frac{1}{2}\cdot \overrightarrow{AC}[/tex]
Which proofs the given affirmation.
