91Ó°ÊÓ

If a line passes through vector\({\bf{a}}\)and is parallel to vector b,then the parametric equation of the line is represented as\({\bf{x}} = {\bf{a}} + t{\bf{b}}\), where\(t\)is a parameter.

Here, \({\bf{x}}\) is represented as \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\).

Consider the parametric equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

Substitute the vectors\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\) in the equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\)as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\)

Thus, \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\).

Substitute \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\) in the above equation.

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\)

Simplify further.

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7t}\\{8t}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 7t}\\{ - 4 + 8t}\end{array}} \right]\end{array}\)

Equate the vectors \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 7t}\\{ - 4 + 8t}\end{array}} \right]\) to obtain the parametric equations.

Thus, the parametric equations are \({x_1} = 3 - 7t\) and \({x_2} = - 4 + 8t\).