Question

In ΔABC, M is the midpoint of

Answer 1

Answer:

AM = MB = MC as $\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \|\overrightarrow{MC}\|$.

Step-by-step explanation:

Let $A(x,y) = (7,1)$, $B(x,y) = (1,-7)$ and $C(x,y) = (1,1)$ vertices of triangle ABC and M is the midpoint of AB. From Linear Algebra and Analytical Geometry, we know that midpoint is represented by the following expression:

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

$M(x,y) = \frac{1}{2}\cdot (7,1)+\frac{1}{2}\cdot (1,-7)$

$M(x,y) = \left(\frac{7}{2},\frac{1}{2}\right)+\left(\frac{1}{2},-\frac{7}{2} \right)$

$M(x,y) = \left(4,-3\right)$

Now, we proceed to calculate $\overrightarrow{AM}$, $\overrightarrow{MB}$ and $\overrightarrow{MC}$ by vector differences:

$\overrightarrow{AM} = M(x,y)-A(x,y)$ (Eq. 1)

$\overrightarrow{AM} = (4,-3)-(7,1)$

$\overrightarrow{AM} = (-3,-4)$

$\overrightarrow {MB} = B(x,y)-M(x,y)$ (Eq. 2)

$\overrightarrow {MB} = (1,-7)-(4,-3)$

$\overrightarrow {MB} = (-3,-4)$

$\overrightarrow {MC} = C(x,y)-M(x,y)$

$\overrightarrow {MC} = (1,1)-(4,-3)$

$\overrightarrow{MC}=(-3,4)$

By Pythagorean Theorem, we get the distances of each relative vector herein:

$\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \sqrt{(-3)^{2}+(-4)^{2}}$

$\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = 5$

$\|\overrightarrow{MC}\| = \sqrt{(-3)^{2}+4^{2}}$

$\|\overrightarrow{MC}\| = 5$

Which proofs that AM = MB = MC. AM = MB = MC as $\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \|\overrightarrow{MC}\|$.