What is an equation of the line that passes through the point (1,6)(1,6) and is perpendicular to the line x+3y=27x+3y=27?

Answer 1

PROBLEM:

What is an equation of the line that passes through the point (1,6) and is perpendicular to

the line x + 3y = 27

STEP 1:

solve the given equation for y, so that it is in the standard Slope-Intercept form: y = mx + b,

where you can easily identify the line's slope

x + 3y = 27

y = - 1 x + 9

3

STEP 2:

because the line you want is to be perpendicular to the provided line, this means you need the negative reciprocal of the slope of given line: m_⊥ = - 1

m

m = _- 1 therefore m_⊥ = _{- ( -} 3 ₎ = 3

3 1

STEP 3:

use the perpendicular slope, m_⊥ and the provided point, (x₁, y₁) and plug everything into

the Point-Slope form of a linear function: y - y₁ = m_⊥(x - x₁)

ANSWER

y - 6 = 3(x - 1) or y = 3(x - 1) + 6

This is the point-slope form of an equation of the line that passes through the point (1,6)

and is perpendicular to the line x + 3y = 27

STEP 4: (optional)

solve the point-slope form of the equation for y and simplify. This will turn the point-slope form into the slope-intercept form of the same line

ANSWER

y = 3x + 3

this is the slope-intercept form of an equation of the line that passes through the point (1,6)

and is perpendicular to the line x + 3y = 27

Answer 2

Hi Casandra!

I see you're looking for an equation of a line. I can help with that.

Your problem does not say what form the equation of the line should be in--there are a few ways to write the equation of a line. Point slope form is the easiest to use for this equation.

(y-y₁) = m(x-x₁) is the point-slope way of writing a linear equation. m represents slope, and x₁ and y₁ come from the point that the line passes through.

So for your new line, the coordinates of your x₁ and y₁ values will come from the point (1,6). x₁ will be 1 and y₁ will be 6. So fill that in to your equation:

(y-6) = m(x-1)

But what about the slope? No worries! You're told that the new line is perpendicular (meaning crossing at a 90 degree angle) to the old line.

A perpendicular line's slope is always the negative reciprocal of the slope of the line it is perpendicular to. So if one line has slope m the perpendicular line's slope is is -(1/m).

So if we find the slope of the old line by...

x+3y=27 --> re-arrange the equation into slope intercept form by...

Subtract X from both sides, canceling out the x's on the left and getting...

3y= -x + 27

Dividing both sides by 3 (remembering to distribute that division by three to both the -x and the 27 on the right side) and getting....

Y = (-1/3)x + 9

So the original line's slope was (-1/3). Therefore using the rule that the perpendicular line's slope will be the negative reciprocal: M_{perpendicular} = -(1/m_original) gives us that:

Old line's slope -1/3

So the new slope will be: positive 3/1 or simply 3.

Remember that equation of the line we already had where we just needed the new m?

Now we have new m=3!

(y-6) = m(x-1) becomes...

(y-6) = 3(x-1)

And that is the equation of a line

that passes through the point (1,6) and is perpendicular to the line x+3y=27.

Have a great day, and let me know if you want any one on one tutoring.

Cheers!

Laura

P.S. If you're looking at an answer key and you see y=3x+3 as the answer...that's right too! It's just written in slope intercept form. Good luck with your work! Feel free to let me know if you ever want some live tutoring.

Answer 3

So our goal is to find a line that goes through the point (1,6) and is perpendicular to x+3y=27.

1. Find the slope
2. We will take x+3y=27 and solve for y. Start by subtracting x from both sides to get 3y=27-x. Then divide both sides by 3 to get y=9-(1/3)x.
3. Notice the slope of the line y = 9 - 1/3x. The slope is -1/3. We must take the negative reciprocal (multiply the slope by flipping it and making it negative) of this in order to find the slope of the line perpendicular to this formula's slope. The negative reciprocal is 3. So the slope of the line we are trying to find is 3.
4. Find the equation
5. Take the point (1,6) and the slope and plug it into the point-slope formula: (y-y₁)=m(x-x₁)
6. Start by labeling the numbers you have so far. The 1 in the point (1,6) is going to be x₁ and the 6 will be y_1. The slope we found earlier was 3. So m will be 3. Plug the numbers into the formula to get (y-6)=3(x-1)
7. Check for double negatives here. Remember that subtracting a negative is the same as adding a positive. Usually what I do on my paper homework is I'll draw a big plus sign over the top of two negatives just so I don't forget. In this problem, we don't have any double negatives, so we can move on.
8. Now we want to distribute the slope into the parentheses on the right side of the equation. 3 times x is 3x and 3 times -1 is -3. So we have (y-6)=3x-3.
9. Lastly, solve for y. Add 6 to both sides to get y by itself on the left. Then we have y=3x+3. This is our answer - y = 3x+3