Finding Point-Slope form of a line that is Perpendicular to One Given! Please help!!

[jwolfe890]

Hello, I have this questions:

Given a line with equation 3x - y = 4. Write an equation in point-slope form of the line perpendicular to the given line through the point (3, 1).

So far, I've been able to reduce the equation to slope intercept form y = 3x – 4, but I am very confused on how to write an equation in point-slope form of the line perpendicular to the given line through the point (3, 1).

If you could offer any advice or insight I would be greatly appreciative!! Thank you!

 

[MarkFL]

When two lines are perpendicular, the product of their slopes is -1. So, what must the slope be of the line perpendicular to:

\(\displaystyle y=3x-4\)

 

[jwolfe890]

-3x

and then the values from that line are (3, 1) and i would just need to add them into point-slope form?

thank you!

 

[MarkFL]

-3x

and then the values from that line are (3, 1) and i would just need to add them into point-slope form?

thank you!

No, that slope isn't correct. When given the line:

\(\displaystyle y=3x-4\)

We see that the slope of this line is 3. So, for the line perpendicular to this one, whose slope we'll call \(\displaystyle m\), we require:

\(\displaystyle 3m=-1\)

So, what is \(\displaystyle m\)?

 

[jwolfe890]

-1/3 after dividing by 3 on both sides

 

[MarkFL]

-1/3 after dividing by 3 on both sides

Yes, correct!

So, now you know the slope of the line, and a point through which the line must pass, so use the point-slope formula to get the equation of the required line.

 

[jwolfe890]

y – 1 = -1/3(x – 3)

y = -1/3x + 1

bam!

thank you so much!

 

[MarkFL]

y – 1 = -1/3(x – 3)

This is correct...

\(\displaystyle y-1=-\frac{1}{3}(x-3)\)

y = -1/3x + 1

Careful! A next step could be:

\(\displaystyle y-1=-\frac{1}{3}x+1\)

Now, add 1 to both sides to get:

\(\displaystyle y=-\frac{1}{3}x+2\)