In Exercises 21 and 22, find a parametric equation of the line \(M\) through \({\mathop{\rm p}\nolimits} \) and \({\mathop{\rm q}\nolimits} \). [Hint: \(M\) is parallel to the vector \({\mathop{\rm q}\nolimits} - p\). See the figure below.]

21.\({\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right]\),\(q = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right]\)

Line \(M\) passing through \(p\) and \(q\) is parallel to the vector \({\mathop{\rm q}\nolimits} - {\mathop{\rm p}\nolimits} \).

It is given that\({\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right]\)and\({\mathop{\rm q}\nolimits} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right]\).

\[\begin{array}{c}q - p = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3 - 2}\\{1 + 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}\\6\end{array}} \right]\end{array}\]

It is known that the equation of the linethrough\(p\)parallel to\({\mathop{\rm v}\nolimits} \)is represented by

\(x = {\mathop{\rm p}\nolimits} + t{\mathop{\rm v}\nolimits} \left( {t\,{\mathop{\rm in}\nolimits} \,\mathbb{R}} \right)\). The solution set of \(Ax = {\mathop{\rm b}\nolimits} \) is a line through \({\mathop{\rm p}\nolimits} \) parallel to the solution set \(Ax = 0\).

Write the parametric equation of line\(M\)parallel to vector\({\mathop{\rm q}\nolimits} - p\)as:

\(\begin{array}{c}x = p + t\left( {q - p} \right)\\ = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 5}\\6\end{array}} \right]\end{array}\)

Thus, the parametric equation of line \(M\) through \(p\) and \(q\) is \(x = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 5}\\6\end{array}} \right]\).