Question

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

Answer 1

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.