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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