Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1). y = x − 3 y = x + 17 y = −3x − 3 y = −3x + 17
First, set it to slope-intercept form. so y=-3x+7 The slope here is -3, and if you want a perpendicular line, you have to change the slope to its opposite reciprocal. The new slope will be 1/3. Plug in 6 for x and -1 for y and then solve for b. -1=(1/3)6 +b -1=2+b -3=b y=(1/3)x-3. All those answers that you have above seem incorrect.... unless I made a mistake. If I did make a mistake, then you can follow the steps I used to try to solve it yourself :)
I plugged it in like you did there and got -1
Did you get -1 for b?
yes but the first two answers are y =1/3 x − 3 y =1/3 x + 17
Ok so I didn't make a mistake. It is y=1/3x-3 :) But how did you get -1 for b?
ok thank you :) and I got it by doing -1=(1/3)6
(1/3) times 6 is 2. you have to -1 by 2 so that gives you -3=b
oh okay would you be able to help me with a few more?
Sure :)
Write the equation of the line that is perpendicular to the line y = x + 4 and passes through the point (−6, 3). y = 3/2x − 1 y =3/2 x + 3 y = −2/3x − 1 y = −2/3x + 3
First, what is the opposite reciprocal of the slope in the given equation?
-4
Umm it's not -4. What is the slope of the equation?
is it 0
No, you have to pretend that there is an imaginary 1 in front of the x. So the slope is 1. The reciprocal of 1 is just 1. The opposite of 1 is -1. So what will the new slope be?
It would be -1
That is correct. Now plug in -1, -6, and 3 and solve for b.
I got 9
b=-3. I think you just made a simple mistake by adding 3 and 6, instead of subtracting.|dw:1342226480711:dw|
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