Find the slope of a line parallel to y=-4x+7

caritatsjq
2022-10-21
Answered

Find the slope of a line parallel to y=-4x+7

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hanfydded1c

Answered 2022-10-22
Author has **17** answers

Your equation is in the Slope-Intercept form:

y=mx+c

where m is the slope; in your case slope is m=−4.

The parallel must have the same slope, i.e., −4.

y=mx+c

where m is the slope; in your case slope is m=−4.

The parallel must have the same slope, i.e., −4.

Tessa Peters

Answered 2022-10-23
Author has **4** answers

This equation is in the form y=mx+c

The gradient is -4.

Perpendicular gradients are negative reciprocals of each other.

${m}_{1}\times {m}_{\_}2=-1$

The negative reciprocal of -4 is $\frac{1}{4}$

This is the required gradient.

Check: $-4\times \frac{1}{4}=-1$

The gradient is -4.

Perpendicular gradients are negative reciprocals of each other.

${m}_{1}\times {m}_{\_}2=-1$

The negative reciprocal of -4 is $\frac{1}{4}$

This is the required gradient.

Check: $-4\times \frac{1}{4}=-1$

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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.

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