Answer:
AM = MB = MC as [tex]\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \|\overrightarrow{MC}\|[/tex].
Step-by-step explanation:
Let [tex]A(x,y) = (7,1)[/tex], [tex]B(x,y) = (1,-7)[/tex] and [tex]C(x,y) = (1,1)[/tex] 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:
[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 (7,1)+\frac{1}{2}\cdot (1,-7)[/tex]
[tex]M(x,y) = \left(\frac{7}{2},\frac{1}{2}\right)+\left(\frac{1}{2},-\frac{7}{2} \right)[/tex]
[tex]M(x,y) = \left(4,-3\right)[/tex]
Now, we proceed to calculate [tex]\overrightarrow{AM}[/tex], [tex]\overrightarrow{MB}[/tex] and [tex]\overrightarrow{MC}[/tex] by vector differences:
[tex]\overrightarrow{AM} = M(x,y)-A(x,y)[/tex] (Eq. 1)
[tex]\overrightarrow{AM} = (4,-3)-(7,1)[/tex]
[tex]\overrightarrow{AM} = (-3,-4)[/tex]
[tex]\overrightarrow {MB} = B(x,y)-M(x,y)[/tex] (Eq. 2)
[tex]\overrightarrow {MB} = (1,-7)-(4,-3)[/tex]
[tex]\overrightarrow {MB} = (-3,-4)[/tex]
[tex]\overrightarrow {MC} = C(x,y)-M(x,y)[/tex]
[tex]\overrightarrow {MC} = (1,1)-(4,-3)[/tex]
[tex]\overrightarrow{MC}=(-3,4)[/tex]
By Pythagorean Theorem, we get the distances of each relative vector herein:
[tex]\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \sqrt{(-3)^{2}+(-4)^{2}}[/tex]
[tex]\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = 5[/tex]
[tex]\|\overrightarrow{MC}\| = \sqrt{(-3)^{2}+4^{2}}[/tex]
[tex]\|\overrightarrow{MC}\| = 5[/tex]
Which proofs that AM = MB = MC. AM = MB = MC as [tex]\|\overrightarrow{AM}\| = \|\overrightarrow{MB}\| = \|\overrightarrow{MC}\|[/tex].
