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find the equation of the line perpendicular to the linear equation fx+gy=p that goes through the point (r,s) How do i do this problem??plz provide a answer. i kno how to get a perpendicualr line but i dont kno how to include points

2006-12-21 14:57:40 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

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 · answer #1 · answered by raj 7 · 0 0

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 · answer #2 · answered by Puggy 7 · 0 0

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 · answer #3 · answered by miss_ooO 2 · 0 0

none bit

2006-12-21 22:59:50 · answer #4 · answered by Anonymous · 0 0

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