Determine the equation of the line passing through (2,6) That is perpendicular to y=−3x+1

tikaj1x
2022-10-18
Answered

Determine the equation of the line passing through (2,6) That is perpendicular to y=−3x+1

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Reese Hobbs

Answered 2022-10-19
Author has **13** answers

It is in the form, y=mx+c where m is the slope & c the y-intercept.

$\therefore m=-3$

Slope of perpendicular line $m}_{1}=-\frac{1}{m}=\frac{1}{3$

Equation of perpendicular line passing thru (2,6) is

$(y-{y}_{1})=m\cdot (x-{x}_{1})$

$y-6=\left(\frac{1}{3}\right)\cdot (x-2)$

$3y-18=x-2$

$3y=x+16$

$\therefore m=-3$

Slope of perpendicular line $m}_{1}=-\frac{1}{m}=\frac{1}{3$

Equation of perpendicular line passing thru (2,6) is

$(y-{y}_{1})=m\cdot (x-{x}_{1})$

$y-6=\left(\frac{1}{3}\right)\cdot (x-2)$

$3y-18=x-2$

$3y=x+16$

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$\left|\begin{array}{ccc}1& -2& 5\\ 4& -5& 8\\ -3& 3& -3\end{array}\right|\left|\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right|=\left|\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right|$

I need to determine the values of the $b$ constants that would guarantee that the linear system is consistent. I tried to find the inverse of the matrix on the left hand side so that I could try and solve for the $x$ variables and see if there are any values of $b$ that would cause the system to be inconsistent but the matrix is singular. I then went on to put the matrix in reduced row echelon form.

$\left|\begin{array}{ccc}1& 0& -3\\ 0& 1& -4\\ 0& 0& 0\end{array}\right|$

From this I was able to derive the equations: ${x}_{1}-3{x}_{3}={b}_{1}$, ${x}_{2}-4{x}_{3}={b}_{2}$, and $0={b}_{3}$. I know that the answer is ${b}_{1}={b}_{2}+{b}_{3}$, but I don't know how to get that with the given information.

I need to determine the values of the $b$ constants that would guarantee that the linear system is consistent. I tried to find the inverse of the matrix on the left hand side so that I could try and solve for the $x$ variables and see if there are any values of $b$ that would cause the system to be inconsistent but the matrix is singular. I then went on to put the matrix in reduced row echelon form.

$\left|\begin{array}{ccc}1& 0& -3\\ 0& 1& -4\\ 0& 0& 0\end{array}\right|$

From this I was able to derive the equations: ${x}_{1}-3{x}_{3}={b}_{1}$, ${x}_{2}-4{x}_{3}={b}_{2}$, and $0={b}_{3}$. I know that the answer is ${b}_{1}={b}_{2}+{b}_{3}$, but I don't know how to get that with the given information.