Find the slope that is perpendicular to the line 2x−7y=−12

dannigurl21ck2
2022-11-21
Answered

Find the slope that is perpendicular to the line 2x−7y=−12

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Cullen Petersen

Answered 2022-11-22
Author has **13** answers

First put the equation in slope-intercept form y=mx+b:

Add −2x to both sides: −7y=−2x−12

Divide everything by $-7:\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}y=\frac{2}{7}x+\frac{12}{7}$

The slope of the given line is $\frac{2}{7}$

The slope of the perpendicular line is the negative reciprocal of the given line: $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}-\frac{7}{2}$

Add −2x to both sides: −7y=−2x−12

Divide everything by $-7:\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}y=\frac{2}{7}x+\frac{12}{7}$

The slope of the given line is $\frac{2}{7}$

The slope of the perpendicular line is the negative reciprocal of the given line: $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}-\frac{7}{2}$

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I have a system of linear equations:

$x-y+2z-t=1$

$2x-3y-z+t=-1$

$x+(\alpha -4)z=\alpha -3$

I have already found that this system has a solution for any value of $\alpha $. Now I need to find the $\alpha $ for which the matrix of the system has a rank=2. The row echelon form of the matrix looks like this:

$\left[\begin{array}{ccccccccc}1& & -1& & 2& & -1& & 1\\ 0& & -1& & -5& & 3& & -3\\ 0& & 0& & \alpha -11& & 4& & \alpha -7\end{array}\right]$

I don't think that the rank of this matrix could be 2 for any value of $\alpha $ but the problem specifically asks for me to prove that it can. Maybe I'm missing something. Any help is appreciated.

$x-y+2z-t=1$

$2x-3y-z+t=-1$

$x+(\alpha -4)z=\alpha -3$

I have already found that this system has a solution for any value of $\alpha $. Now I need to find the $\alpha $ for which the matrix of the system has a rank=2. The row echelon form of the matrix looks like this:

$\left[\begin{array}{ccccccccc}1& & -1& & 2& & -1& & 1\\ 0& & -1& & -5& & 3& & -3\\ 0& & 0& & \alpha -11& & 4& & \alpha -7\end{array}\right]$

I don't think that the rank of this matrix could be 2 for any value of $\alpha $ but the problem specifically asks for me to prove that it can. Maybe I'm missing something. Any help is appreciated.

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If not, is there an analytical solution for some particular form of $\omega (t)$?

If the answer is again no, what software would be recommendable for numerical solution?

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with $x(t)$ being a real function (and $\omega (t)$ being also time-dependent).

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If not, is there an analytical solution for some particular form of $\omega (t)$?

If the answer is again no, what software would be recommendable for numerical solution?

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Background

It is relatively easier to find a solution to a system of linear equations in the form of $A\mathbf{\text{v}}=\mathbf{\text{b}}$ given the matrix $A$. But what systematic ways are there that allows us to obtain a matrix given a equation?

For example, consider the following equations with all terms existing in $\mathbb{R}$

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Although it is easy to see that $a=\frac{1}{2},e=\frac{1}{3},i=\frac{1}{4}$ with all other terms being 0 is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = k?

Why I am interested in such question

Consider the vector space ${P}_{2}(\mathbb{R})$, the problem of finding a basis $\beta $ such that $[{x}^{2}+x+1{]}_{\beta}=(2,3,4{)}^{T}$ can be reduced to a problem that has been stated above.

It is relatively easier to find a solution to a system of linear equations in the form of $A\mathbf{\text{v}}=\mathbf{\text{b}}$ given the matrix $A$. But what systematic ways are there that allows us to obtain a matrix given a equation?

For example, consider the following equations with all terms existing in $\mathbb{R}$

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Although it is easy to see that $a=\frac{1}{2},e=\frac{1}{3},i=\frac{1}{4}$ with all other terms being 0 is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = k?

Why I am interested in such question

Consider the vector space ${P}_{2}(\mathbb{R})$, the problem of finding a basis $\beta $ such that $[{x}^{2}+x+1{]}_{\beta}=(2,3,4{)}^{T}$ can be reduced to a problem that has been stated above.