line perpendicular to fx+gy=p
slope of the given line
gy=-fx+p
slope=-f/g
slope of the required line=g/f
passes through (r,s)
the equation
y-s=g/f(x-r)
f(y-s)=g(x-r)
fy-fs-gx+gr=0
fy-gx=fs-gr
2006-12-21 15:05:41
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answer #1
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answered by raj 7
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fx + gy = p.
Your first step is to calculate the slope, and to put the above equation in slope-intercept form. This would be of the form
"y = mx + b".
gy = -fx + p
y = (-f/g)x + p/g
Therefore, the slope m is equal to -f/g.
To obtain the perpendicular line, what you have to associate with "perpendicular" is "negative reciprocal". To elaborate, the perpendicular line is going to have a negative reciprocal slope. So for m = -f/g, take the negative reciprocal.
This is relatively simple to do. First, flip the terms (making it -g/f), and then take the negative of it (effectively removing the minus sign, for an answer of (g/f)
Therefore, the equation of our perpendicular line is going to look like
y = (g/f)x + b
But this is incomplete, because we need to solve for b. What we do now is recall that the slope of a line is equal to
m = (y2 - y1) / (x2 - x1)
However, we just solved for the slope m (it's equal to g/f). What we do now, to obtain the equation of the line is to plug in the values (r,s) for (x1,y1) {since the lines goes through the point (r,s) and (x,y) for (x2,y2) {since (x,y) represents the general term of the line).
g/f = (y - s) / (x - r)
Multiplying both sides by (x - r), we get
(g/f) (x - r) = y - s
Distributing (g/f) on the left hand side,
(g/f)x - (gr)/f = y - s
And bringing the -s over to the left hand side yields
(g/f)x - (gr)/f + s = y, OR
y = (g/f)x + {s - (gr)/f}
I put the second part in { } brackets to make it clear what the y-intercept is.
2006-12-21 23:12:03
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answer #2
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answered by Puggy 7
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use this formula
y-y1=m(x-x1)
gradient for perpendicular line use this formula
m x m1 = -1
2006-12-21 23:04:19
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answer #3
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answered by miss_ooO 2
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