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

jwolfe890

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

y=3x4\displaystyle y=3x-4
 
-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!
 
-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:

y=3x4\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 m\displaystyle m, we require:

3m=1\displaystyle 3m=-1

So, what is m\displaystyle m?
 
-1/3 after dividing by 3 on both sides

Yes, correct! :D

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. :D
 
y – 1 = -1/3(x – 3)

This is correct...

y1=13(x3)\displaystyle y-1=-\frac{1}{3}(x-3)

y = -1/3x + 1

Careful! A next step could be:

y1=13x+1\displaystyle y-1=-\frac{1}{3}x+1

Now, add 1 to both sides to get:

y=13x+2\displaystyle y=-\frac{1}{3}x+2
 
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