The midpoint is the average of the endpoints.

\(\displaystyle (\frac{3-1}{2}, \frac{11-7}{2})=(\frac{2}{2},\frac{4}{2})=(1,2)\)

Use the midpoint formula, \(\displaystyle mid=(\frac{x_{1}+x_2}{2}, \frac{y_1+y_2}{2})\), with our points  \(\displaystyle (-4,6)\) and \(\displaystyle (8,2)\).

\(\displaystyle mid=(\frac{-4+8}{2},\frac{6+2}{2})\)

\(\displaystyle mid=(\frac{4}{2},\frac{8}{2})=(2,4)\)

To find the midpoint of a line segment, use the standard midpoint equation: 

\(\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\)

Plug in the given points and simplify:

\(\displaystyle (\frac{-12+4}{2},\frac{-1+3}{2})=(-4,1)\)

To find the midpoint of a line segment, we find the average of the \(\displaystyle x\)-values and the average of the \(\displaystyle y\)-values. Thus, we have that: 

Midpoint = \(\displaystyle \left(\frac{-1 + 5}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{4}{2}, \frac{2}{2}\right) = (2, 1)\)

The formula for the length of a line is very similiar to the pythagorean theorem:

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

Plug in our given numbers to solve:

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

\(\displaystyle (5-0)^2+(15-3)^2=l^2\)

\(\displaystyle 5^2+12^2=l^2\)

\(\displaystyle 25+144=l^2\)

\(\displaystyle 169=l^2\)

\(\displaystyle \sqrt{169}=\sqrt{l^2}\)

\(\displaystyle 13=l\)

If B is the midpoint of AC, then AC is twice as long as AB. We are told that AB=6.

\(\displaystyle A------B------C\)

The diagram shows six units between points A and B, with B as the midpoint of segment AC. Therefore segment BC is also six units long, so line AC is twelve units long.

\(\displaystyle \small (2)(6)=12\)

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

Plug in the given values and solve for the length.

\(\displaystyle (5-0)^2+(7-1)^2=l^2\)

\(\displaystyle (5)^2+(6)^2=l^2\)

\(\displaystyle 25+36=l^2\)

\(\displaystyle 61=l^2\)

\(\displaystyle \sqrt{61}=\sqrt{l^2}\)

 \(\displaystyle \sqrt{61}=l\)

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

Plug in the given values and solve for the length.

\(\displaystyle (9-1)^2+(9-6)^2=l^2\)

\(\displaystyle (8)^2+(3)^2=l^2\)

\(\displaystyle 64+9=l^2\)

\(\displaystyle 73=l^2\)

\(\displaystyle \sqrt{73}=\sqrt{l^2}\)

\(\displaystyle \sqrt{73}=l\)

Find the slope between the two points:

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}\)

Next, use the slope-intercept form of the equation:

\(\displaystyle y=mx+b\) or \(\displaystyle 15=\frac{5}{3}(5)+b\) where \(\displaystyle b=\frac{20}{3}\)

So the equation becomes \(\displaystyle y=\frac{5}{3}x+ \frac{20}{3}\) or in standard form \(\displaystyle -5x+3y=20\).

The formula for the length of a line is very similiar to the pythagorean theorem:

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

Plug in our given numbers to solve:

\(\displaystyle (x_2-x_1)^2+(y_2-y_1)^2=l^2\)

\(\displaystyle (8-4)^2+(13-10)^2=l^2\)

\(\displaystyle 4^2+3^2=l^2\)

\(\displaystyle 16+9=l^2\)

\(\displaystyle 25=l^2\)

\(\displaystyle \sqrt{25}=\sqrt{l^2}\)

\(\displaystyle 5=l\)