Question

NO LINKS!! URGENT HELP PLEASE!!

Please help me with #20​

Answer 1

`\text{Midpoint of $AC$}=\left(-2,2\right)`

`\text{Midpoint of $AB$}=\left((5)/(2),3\right)`

`\text{Slope of midsegment}=(2)/(9)`

`\text{Slope of $AC$}=3`

`\text{Slope of $BC$}=(2)/(9)`

`\text{Length of midsegment}=\frac{√(85)}2`

`\text{Length of $BC$}=√(85)`

Explanation:

Given points:

• A = (-1, 5)
• B = (6, 1)
• C = (-3, -1)

To determine the midpoints of AC and AB, substitute the given points into the midpoint formula.

`\boxed{\begin{minipage}{7.4 cm}\underline{Midpoint between two points}\\\\Midpoint $=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints.\\\end{minipage}}`

`\begin{aligned}\text{Midpoint of $AC$}&=\left((x_C+x_A)/(2),(y_C+y_A)/(2)\right)\\\\&=\left((-3-1)/(2),(-1+5)/(2)\right)\\\\&=\left((-4)/(2),(4)/(2)\right)\\\\&=\left(-2,2\right)\end{aligned}`

`\begin{aligned}\text{Midpoint of $AB$}&=\left((x_B+x_A)/(2),(y_B+y_A)/(2)\right)\\\\&=\left((6-1)/(2),(1+5)/(2)\right)\\\\&=\left((5)/(2),(6)/(2)\right)\\\\&=\left((5)/(2),3\right)\end{aligned}`

`\hrulefill`

`\boxed{\begin{minipage}{8cm}\underline{Slope Formula}\\\\Slope $(m)=(y_2-y_1)/(x_2-x_1)$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line.\\\end{minipage}}`

To determine the slope of the midsegment, substitute the midpoints of AC and AB into the slope formula:

`\begin{aligned}\text{Slope of midsegment}&=(y_(AB)-y_(AC))/(x_(AB)-x_(AC))\\\\&=(2-3)/(-2-(5)/(2))\\\\&=(-1)/(-(9)/(2))\\\\&=(2)/(9)\end{aligned}`

Therefore, the slope of the midsegment is 2/9.

To find the slope of AC, substitute the points A and C into the slope formula:

`\begin{aligned}\text{Slope of $AC$}&=(y_(C)-y_(A))/(x_(C)-x_(A))\\\\&=(-1-5)/(-3-(-1))\\\\&=(-6)/(-2)\\\\&=3\end{aligned}`

Therefore, the slope of the AC is 3.

**Note** There may be an error in the question. I think you are supposed to find the slope of BC (not AC) since there is no relationship between the slopes of the midsegment and AC, but there is a relationship between the slopes of the midsegment and BC.

`\begin{aligned}\text{Slope of $BC$}&=(y_(C)-y_(B))/(x_(C)-x_(B))\\\\&=(-1-1)/(-3-6)\\\\&=(-2)/(-9)\\\\&=(2)/(9)\end{aligned}`

The slope of BC is 2/9, so the slopes of the midsegment and BC are the same. This implies that the midsegment and BC are parallel.

`\hrulefill`

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance Formula}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where:\\ \phantom{ww}$\bullet$ $d$ is the distance between two points. \\\phantom{ww}$\bullet$ $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

To find the length of the midsegment, substitute the endpoints (-2, 2) and (5/2, 3) into the distance formula:

`\begin{aligned}\text{Length of midsegment}&=\sqrt{\left((5)/(2)-(-2)\right)^2+(3-2)^2}\\\\&=\sqrt{\left((9)/(2)\right)^2+(1)^2}\\\\&=\sqrt{(81)/(4)+1}\\\\&=\sqrt{(85)/(4)}\\\\&=\frac{√(85)}2\end{aligned}`

To find the length of the BC, substitute points B and C into the distance formula:

`\begin{aligned}\text{Length of $BC$}&=√((x_C-x_B)^2+(y_C-y_B)^2)\\&=√((-3-6)^2+(-1-1)^2)\\&=√((-9)^2+(-2)^2)\\&=√(81+4)\\&=√(85)\end{aligned}`

Therefore, the length of BC is twice the length of the midsegment.