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#1 2005-12-05 01:44:51
- RickyOswaldIOW
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- Registered: 2005-11-18
- Posts: 212
Perpendicular line equations.
I've completed a few of these with no problem but I am stuck on:
2x + 3y + 2 = 0, find the equation of the perpendicular line passing through (4, 7).
2x + 3y + 2 = 0
3y = -2x - 2
y = (-2x -2)/3
perpendicular line would be :
y = 3/2 x + 4 1/3?
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#2 2005-12-05 01:58:08
- RickyOswaldIOW
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- Registered: 2005-11-18
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Re: Perpendicular line equations.
The perpendicular line would be:
y = 3/2 x + 1
2y = 3x + 2
What is the equation used to get to this number?
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#3 2005-12-05 03:26:06
- mathsyperson
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- Registered: 2005-06-22
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Re: Perpendicular line equations.
You got the gradient right, because to find a perpendicular gradient, you divide -1 by the original gradient.
If you know a line has gradient g and passes through (x,y), then its y-intercept is y - (g*x).
For your one, it passes through (4,7) and has a gradient of 3/2, so the y-intercept is 7 - (3/2*4) = 1.
Why did the vector cross the road?
It wanted to be normal.
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#4 2005-12-05 03:27:07
- RickyOswaldIOW
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Re: Perpendicular line equations.
Thankyou mathsy!
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#5 2005-12-05 03:30:55
- irspow
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- Registered: 2005-11-24
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Re: Perpendicular line equations.
Perpendicular lines slopes always share a negative inverse relationship.
Mathematically this means that the product of the two slopes always equals negative one.
m1(m2) = -1
A linear equation can always be written as mx + b = y
You were on the right track to solving this problem.
Solve for y for the original equation: y= -2/3 x -2/3
Therefore the slope is -2/3
If the second line is perpendicular then -2/3 m2 = -1, so m2 = 3/2 or 1.5
The second equation must then be y = 1.5x + b
Plugging in the point specified by your question: 7 = 1.5(4) + b
Solving for b and keeping the slope give you your perpendicular line equation
y = 1.5x + 1
Last edited by irspow (2005-12-05 03:32:22)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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#6 2005-12-05 03:52:09
- RickyOswaldIOW
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- Registered: 2005-11-18
- Posts: 212
Re: Perpendicular line equations.
The perpendicular line would be:
y = 3/2 x + 1
2y = 3x + 2
Thanks irspow but you're a bit late 
Aloha Nui means Goodbye.
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#7 2005-12-05 04:47:29
- irspow
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- Registered: 2005-11-24
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Re: Perpendicular line equations.
Sorry, my fingers are slower than my head.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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