Question

find the equation of the line that passes through P(7,20) and Q the midpoint of R (-3,5) and S (5,11)

Answer 1

$R (-3,5), \ \ \ S (5,11) \ midpoint \ of \ R \ and \ S \\ \\ Midpoint \ Formula \\\\(x,y)= \left ( \frac{x_{1}+x_{2}}{2},\frac {{}y_{1}+y_{2}}{2} \right ) \\ \\Q= \left ( \frac {-3+5}{2},\frac { 5+11}{2} \right ) \\ \\Q= \left ( \frac {2}{2},\frac { 16}{2} \right ) \\ \\Q= \left ( 1 ,8) \right )$

$the \ equation \ of \ the \ line \ that \ passes \ through \ P(7,20) \ and \ Q (1,8)\\\\First \ find \ the \ slope \ of \ the \ line \ thru \ the \ points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} } \\ \\m=\frac{ 8-20}{1-7 } =\frac{-12}{-6}=2\\\\the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\\\20=2\cdot 7+b \\\\20=14+b\\\\b=20-14\\b=6\\\\y=2x+6$

Answer 2

First we have to find midpoint of R and S.
We can use formula such for it.$Qx= \frac{Rx+Sx}{2}$ and $Qy= \frac{Ry+Sy}{2}$.
We obtained coordinates of point Q 
$Qx= \frac{-3+5}{2}=1$ and $Qy= \frac{5+11}{2}=8$

Now, we can find the line equation using formula y=ax+b. 
We can substitute coordinates of P and Q to this formula and solving system of equation get the answer.

After substituting we obtaind such system
$\left \{ {{20=7a+b } \atop {8=a+b}} \right.$

From the system of equation we obtain result
$\left \{ {{a=2} \atop {b=6}} \right.$ 

Now we can put our resuts to general line equation.
$y=2x+6$