Chapter 13, Problem 33RP

To determine

To find: The intersection of line MN and line PX such that for $P=\left(-8,1\right)$ , $Q=\left(0,5\right)$ and $S=\left(4,1\right)$ .

Answer to Problem 33RP

Thepoint of intersection is $\left(\frac{-10}{3},\frac{7}{3}\right)$ .

Explanation of Solution

Consider the mid-point between the points between P and Q is obtained as,

  $\begin{array}{c}M=\left(\frac{-8+0}{2},\frac{1+5}{2}\right)\\ =\left(-4,3\right)\end{array}$

Consider the mid-point between the points between P and Sis obtained as,

  $\begin{array}{c}N=\left(\frac{-8-4}{2},\frac{1+1}{2}\right)\\ =\left(-2,1\right)\end{array}$

The equation of the line MN is obtained as,

  $\begin{array}{l}y-1=-\left(x+2\right)\\ y+x=-1\end{array}$

The equation of the line PX is obtained as,

  $\begin{array}{l}y-1=\frac{2}{7}\left(x+8\right)\\ 7y-2x=23\end{array}$

Solving the above equation the point of intersection of the two line obtained is $\left(\frac{-10}{3},\frac{7}{3}\right)$ .