# Determine the parametric, vector and scalar equations for line passing

sklicatias 2021-11-29 Answered
Determine the parametric, vector and scalar equations for line passing through $$\displaystyle{\left(-{2},\ {11}\right)}$$ and $$\displaystyle{\left({6},\ {1}\right)}$$

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## Expert Answer

Leory2000
Answered 2021-11-30 Author has 7703 answers
Step 1
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) $$\displaystyle{y}={11}-{10}$$
$$\displaystyle{y}-{11}=-{10}{t}$$
$$\displaystyle{t}={\frac{{{11}-{y}}}{{{10}}}}$$
$$\displaystyle\Rightarrow{x}=-{2}+{8}{\left({\frac{{{11}-{y}}}{{{10}}}}\right)}$$
$$\displaystyle{x}=-{10}+{44}-{4}{y}$$
$$\displaystyle{x}={34}-{4}{y}$$
$$\displaystyle\Rightarrow{x}+{4}{y}-{34}={0}$$

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