Given:

to find parametric equations through a point \(\displaystyle{\left(-{2},\ {11}\right)}\) and \(\displaystyle{\left({6},\ {1}\right)}\)

Direction vextor \(\displaystyle\vec{{{v}}}={\left\langle{6}+{2},\ {1}-{11}\right\rangle}\)

\(\displaystyle={\left\langle{8},\ -{10}\right\rangle}\)

Parametric equations are:

\(\displaystyle{x}={x}_{{{0}}}+{a}{t},\ {y}={y}_{{{0}}}+{b}{t}\)

\(\displaystyle{x}=-{2}+{8}{t},\ {y}={11}-{10}{t}\)

Step 2

Vector equation:

\(\displaystyle{r}={a}+{t}{b}\)

\[=\left(\begin{matrix} -2 \\ 11 \end{matrix}\right)+t\left(\begin{matrix} 8 \\ -10 \end{matrix}\right)\]

Scalar equation:

1) \(\displaystyle{x}=-{2}+{8}{t}\)

2) \(\displaystyle{y}={11}-{10}\)

\(\displaystyle{y}-{11}=-{10}{t}\)

\(\displaystyle{t}={\frac{{{11}-{y}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{x}=-{2}+{8}{\left({\frac{{{11}-{y}}}{{{10}}}}\right)}\)

\(\displaystyle{x}=-{10}+{44}-{4}{y}\)

\(\displaystyle{x}={34}-{4}{y}\)

\(\displaystyle\Rightarrow{x}+{4}{y}-{34}={0}\)