In Triangle ABC with vertices A(-5,-4), B(3,-2), and C(-1,6), M is the midpoint of AB and N is the midpoint of BC. Show that MN = 1/2 AC. Provide your complete solutions and proofs in your paper homework and enter the numeric answers online

The midpoint of a line divides the line into equal segments.

See below for the proof of [tex]\mathbf{MN = \frac{1}{2}AC}[/tex]

The given parameters are:

[tex]\mathbf{A = (-5,-4)}[/tex]

[tex]\mathbf{B = (3,-2)}[/tex]

[tex]\mathbf{C = (-1,6)}[/tex]

The coordinates of M (midpoint of AB) are calculated as follows:

[tex]\mathbf{M = \frac{1}{2}(x_1 + x_2, y_1+y_2)}[/tex]

So, we have:

[tex]\mathbf{M = \frac{1}{2}(-5 + 3, -4-2)}[/tex]

[tex]\mathbf{M = \frac{1}{2}(-2, -6)}[/tex]

[tex]\mathbf{M = (-1, -3)}[/tex]

The coordinates of N (midpoint of BC) are calculated as follows:

[tex]\mathbf{N = \frac{1}{2}(x_1 + x_2, y_1+y_2)}[/tex]

So, we have:

[tex]\mathbf{N = \frac{1}{2}(3 - 1, -2 +6)}[/tex]

[tex]\mathbf{N = \frac{1}{2}(2, 4)}[/tex]

[tex]\mathbf{N = (1, 2)}[/tex]

Distance MN is:

[tex]\mathbf{MN = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}[/tex]

So, we have:

[tex]\mathbf{MN = \sqrt{(-1 - 1)^2 + (-3 - 2)^2}}[/tex]

[tex]\mathbf{MN = \sqrt{29}}[/tex]

Distance AC is calculated as follows:

[tex]\mathbf{AC= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}[/tex]

So, we have:

[tex]\mathbf{AC= \sqrt{(-5 - -1)^2 + (-4 - 6)^2}}[/tex]

[tex]\mathbf{AC= \sqrt{116}}[/tex]

Split

[tex]\mathbf{AC= \sqrt{4 \times 29}}[/tex]

Take positive square root of 4

[tex]\mathbf{AC= 2\sqrt{29}}[/tex]

To prove that:

[tex]\mathbf{MN = \frac{1}{2}AC}[/tex]

We have:

[tex]\mathbf{\sqrt{29} = \frac{1}{2} \times 2\sqrt{29}}[/tex]

[tex]\mathbf{\sqrt{29} = \sqrt{29}}[/tex]

The above equation is true.

Hence. [tex]\mathbf{MN = \frac{1}{2}AC}[/tex]

Read more about midpoints and distance at:

https://brainly.com/question/11231122

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