Question

What is the equation of a line that is perpendicular to −x+3y=9 and passes through the point (−3, 2) ?

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Answer 1

We are given that

line is perpendicular to $-x+3y=9$

Firstly, we will solve for y

$-x+3y=9$

Add both sides by x

$-x+3y+x=x+9$

$3y=x+9$

Divide both sides by 3

$\frac{3y}{3} =\frac{x}{3} +\frac{9}{3}$

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

now, we can compare it with slope -intercept form of line

$y =mx +b$

where m is slope

so, we can find m

$m=\frac{1}{3}$

now, we are given that line perpendicular

so, we can find slope of perpendicular line

we know that product of two perpendicular lines are always -1

so,

$m_p\times \frac{1}{3} =-1$

now, we can solve for mp

$m_p=-3$

now, it passes through point (−3, 2)

so, $x_1=-3,y_1=2$

now, we can use point slope form of line

$y-y_1=m(x-x_1)$

we can plug values

$y-2=-3(x+3)$

now, we can solve for y

$y-2=-3x-9$

Add both sides by 2

$y-2+2=-3x-9+2$

$y=-3x-7$

So, equation of our line is

$y=-3x-7$.................Answer

Answer 2

The equation of line that is perpendicular to the line $-x+3y =9$  and passes through the point $\left({-3,2}\right)$  is $\boxed{{\mathbf{y=-3x-7}}}$ .

Further explanation:

It is given that the equation of line is $-x+3y =9$  and passes through point $\left({-3,2}\right)$ .

Rewrite the given equation $-x+3y =9$  as follows:

$\begin{aligned}-x+3y&=9\\3y&=9+x\\y&=\frac{9}{3}+\frac{1}{3}x\\y&=\frac{1}{3}x+3\\\end{aligned}$

Now, compare the obtained equation of line $y=\frac{1}{3}x+3$  with the standard equation of line $y=mx+b$ .

$\begin{aligned}m&=\frac{1}{3}\\b&=3\\\end{aligned}$

Therefore, the slope is $\frac{1}{3}$ .

It is given that both lines are perpendicular to each other so the product of slope must be equal to $-1$ .

${m_1}\cdot {m_2}=-1$                                                          …… (1)

Substitute $\frac{1}{3}$  for ${m_1}$  in equation (1) to obtain the value of slope ${m_2}$ .

$\begin{aligned}\frac{1}{3}\cdot{m_2}&=-1\\{m_2}&=-3\\\end{aligned}$

Therefore, the slope is $-3$ .

It is given that the line passes through point $\left({-3,2}\right)$ .

The point-slope form of the equation of a line with slope $m$  passes through point $\left({{x_1},{y_1}}\right)$ is represented as follows:

$y-{y_1}=m\left({x-{x_1}}\right)$                                      …… (2)

Substitute $-3$  for ${x_1}$ , $2$  for ${y_1}$  and $-3$  for $m$  in equation (2) to obtain the equation of line.

$\begin{aligned}y-2&=-3\left({x-\left({-3}\right)}\right)\\y-2&=-3\left({x+3}\right)\\y&=-3x-9+2\\y&=-3x-7\\\end{aligned}$

Therefore, the equation of line is $y=-3x-7$ .

Thus, the equation of line that is perpendicular to the line $-x+3y=9$  and passes through the point $\left({-3,2}\right)$  is $\boxed{{\mathbf{y=-3x-7}}}$ .

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, equation of line, line, passes through point.