Finding $k$ such that $\binom{-2}{k}$ is the direction vector of the line $y=\frac17(4x+1)$ I'm new here, I asked my friend about this question and he told me to go on this math forum. If someone can get me started on these questions it would make my day. Thank You


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*Find the value of $k$ such that $\dbinom{-2}{k}$ is a direction vector of the line with equation $y=\dfrac{4x+1}{7}$.


*Find an equation for the line with vector form $\dbinom{5}{-2}+t\dbinom{2}{6}$ in the form $y=mx+b$.

 A: On a plane, any line can be presented in two ways:

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*A point on the line and the line direction vector, pointing along the line. Like $\mathbf{x}=\mathbf{a}+\mathbf{v}t$.

*A linear equation about the coordinates of a point, belonging to the line -- the equation is true for a point on the line and false everywhere else. These equations looks like $Ax+By+C=0$ or $\mathbf{n}.(\mathbf{x}-\mathbf{a})=0$, where $\mathbf{n}$ is the line normal vector (and perpendicular to the line), $\mathbf{a}$ is a point on the line and $\mathbf{x}$ is any point we're checking whether it's on the line or not.

So here we just have to convert one equation form to another. To convert from 1 to 2 we rewrite $\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x_0\\y_0\end{pmatrix}+\begin{pmatrix}v_x \\v_y\end{pmatrix}t$ as $\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}(=t)$ and convert it to $Ax+By+C=0$ by just clearing the denominator.
To convert from $Ax+By+C=0$ to 1 we can take any vector, perpendicular to $\mathbf{n}=\begin{pmatrix}A \\B\end{pmatrix}$ (e.g. $\begin{pmatrix}-B \\A\end{pmatrix}$) as direction vector and also we have to find a point, belonging to the line, e.g. by intersection with $x=0$ or $y=0$ or any other convinient line.
$\begin{pmatrix}-2 \\k\end{pmatrix}$ is orthogonal to $\begin{pmatrix}4 \\-7\end{pmatrix}$ when $k=-\frac{8}{7}$ and $\frac{x-5}{2}=\frac{y-2}{6}$ is $y=3x-17$.
