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Lines

by vladmire » Tue Jan 06, 2009 7:53 pm
How can I find the equation of the line that contains the point (−1, 3) and which is perpendicular to the line with the equation y + 2x+ 1 = 0.

Is the answer to this y=1/2x + 7/2


Second question (y^10/125y^-5)^-2/3
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Re: Lines

by vivek.kapoor83 » Tue Jan 06, 2009 8:54 pm
vladmire wrote:How can I find the equation of the line that contains the point (−1, 3) and which is perpendicular to the line with the equation y + 2x+ 1 = 0.

Is the answer to this y=1/2x + 7/2


Second question (y^10/125y^-5)^-2/3

Eq of line = y =mx+c
where m is slope
so, from given eq. y = -2x-1
Slope = -2
Now, If 2 lines are perpendicular, product of their slope = -1
Let m1 b the slope of 2nd line

then m1* -2 = -1
m1 =1/2

Now, Eq of line
y =mx+c
(-1,3) is given'
3 = -1/2+C
c = 3+1/2
c =7/2
So, y = 1/2x+7/2should be the ans IMO
Last edited by vivek.kapoor83 on Tue Jan 06, 2009 8:57 pm, edited 1 time in total.
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Re: Lines

by logitech » Tue Jan 06, 2009 8:55 pm
How can I find the equation of the line that contains the point (−1, 3) and which is perpendicular to the line with the equation y + 2x+ 1 = 0.


The slope of the equation is -2, so if you have two slopes that are perpendicilar, their product needs to be -1 so the slope of the equation is 1/2

Y=1/2 x + B

For X=-1 Y must be 3

SO

3 = -1/2 + b

B = 7/2

So y=1/2 x + 7/2

Is the answer to this y=1/2x + 7/2 YESS




Second question (y^10/125y^-5)^-2/3


(y^15/5^3)^-2/3 = y^-10/5^-2 = 25/y^10
LGTCH
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