Question

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

Answer 1

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

See below for the proof of $\mathbf{MN = \frac{1}{2}AC}$

The given parameters are:

$\mathbf{A = (-5,-4)}$

$\mathbf{B = (3,-2)}$

$\mathbf{C = (-1,6)}$

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

$\mathbf{M = \frac{1}{2}(x_1 + x_2, y_1+y_2)}$

So, we have:

$\mathbf{M = \frac{1}{2}(-5 + 3, -4-2)}$

$\mathbf{M = \frac{1}{2}(-2, -6)}$

$\mathbf{M = (-1, -3)}$

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

$\mathbf{N = \frac{1}{2}(x_1 + x_2, y_1+y_2)}$

So, we have:

$\mathbf{N = \frac{1}{2}(3 - 1, -2 +6)}$

$\mathbf{N = \frac{1}{2}(2, 4)}$

$\mathbf{N = (1, 2)}$

Distance MN is:

$\mathbf{MN = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

So, we have:

$\mathbf{MN = \sqrt{(-1 - 1)^2 + (-3 - 2)^2}}$

$\mathbf{MN = \sqrt{29}}$

Distance AC is calculated as follows:

$\mathbf{AC= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

So, we have:

$\mathbf{AC= \sqrt{(-5 - -1)^2 + (-4 - 6)^2}}$

$\mathbf{AC= \sqrt{116}}$

Split

$\mathbf{AC= \sqrt{4 \times 29}}$

Take positive square root of 4

$\mathbf{AC= 2\sqrt{29}}$

To prove that:

$\mathbf{MN = \frac{1}{2}AC}$

We have:

$\mathbf{\sqrt{29} = \frac{1}{2} \times 2\sqrt{29}}$

$\mathbf{\sqrt{29} = \sqrt{29}}$

The above equation is true.

Hence. $\mathbf{MN = \frac{1}{2}AC}$

Read more about midpoints and distance at:

brainly.com/question/11231122