Find an equation of a line containing the point (-2, 2), and parallel to the line 2(y + 1) = x

Alexander Lewis
2022-10-29
Answered

Find an equation of a line containing the point (-2, 2), and parallel to the line 2(y + 1) = x

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encaselatqr

Answered 2022-10-30
Author has **11** answers

We are given a point on the line and have indirectly been given the slope as well, because lines that are parallel have the same slope.

Simplify the equation of the given line.

$2(y+1)=x\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\to y+1=\frac{1}{2}x$

Then: $y=\frac{1}{2}x-1\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ this gives the slope as $m=\frac{1}{2}$

Now use the point slope formula for a line:

$y-{y}_{1}=m(x-{x}_{1})\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ with $(-2,2)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}m=\frac{1}{2}$

$y-2=\frac{1}{2}(x-(-2))$

$y=\frac{1}{2}x+1+2$

$y=\frac{1}{2}x+3$

Simplify the equation of the given line.

$2(y+1)=x\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\to y+1=\frac{1}{2}x$

Then: $y=\frac{1}{2}x-1\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ this gives the slope as $m=\frac{1}{2}$

Now use the point slope formula for a line:

$y-{y}_{1}=m(x-{x}_{1})\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ with $(-2,2)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}m=\frac{1}{2}$

$y-2=\frac{1}{2}(x-(-2))$

$y=\frac{1}{2}x+1+2$

$y=\frac{1}{2}x+3$

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I am given the following matrix A and I need to find a nullspace of this matrix.

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?

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