the co-ordinates of foot of perpendicular drawn from point (3,4)on the line 2x+y-7=0 is?
Equation of line perpendicular to 2x+y-7=0
is of the form x - 2y + c = 0
(Swap coefficients of x and y and change sign between)
As it passes through (3, 4)
3 - 2*4 + c = 0
c = 5
Thus required perpendicular is x - 2y + 5 = 0 .... (1)
Line is 2x + y - 7 = 0 ...(2)
The foot of the perpendicular is where these two lines intersect.
Solving by eliminating y
2 * equation (2) + equation (1)
5x -9 = 0
x = 9/5
Substitute into Equation (1)
9/5 - 2y + 5 = 0
2y = 34/5
So y = 17/5
Check in Equation (2)
2*9/5 + 17/5 - 7
= 18/5 + 17/5 - 7
= 35/5 - 7
= 0 ... YESSSSSSSSSSSSSSS
So the solution is (9/5, 17/5)
2007-05-10 21:25:46
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answer #1
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answered by Wal C 6
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Find the co-ordinates of the foot of the perpendicular drawn from point (3,4) to the line 2x + y - 7 = 0.
This is just the intersection of the given line and the perpendicular line thru the point (3,4).
The given line is
2x + y - 7 = 0
y = -2x + 7
Its slope m = -2
The slope of a perpendicular line is the negative reciprocal.
m' = -1/m = 1/2
Now use the point slope form to write the equation of the perpendicular line thru (3,4).
y - 4 = (1/2)(x - 3)
y = (1/2)x - 3/2 + 4
y = (1/2)x + 5/2
The intersection of the two lines is the foot of the perpendicular.
y = -2x + 7 = (1/2)x + 5/2
9/2 = (5/2)x
x = 9/5
Plug back into the formula of one of the lines to solve for y.
y = (1/2)x + 5/2
y = (1/2)(9/5) + 5/2 = 9/10 + 5/2 = 34/10 = 17/5
The foot of the perpendicular is (x,y) = (9/5, 17/5).
2007-05-10 20:01:27
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answer #2
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answered by Northstar 7
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The equation of a line perpendicular to a given line ax + by using + c = 0 is bx - ay +ok = 0, the place ok is an arbitrary consistent. So the line perpendicular to y=2x+3 i.e. 2x-y+3=0 is x+2y+ok=0 Now because of the fact it passes (4,7) so 2*7+4+ok=0 > 14+4+ok=0- > ok =-18 So the mandatory line x+2y-18=0
2016-12-17 09:49:04
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answer #3
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answered by Anonymous
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the K of the line 2x+y-7=0 is ,K(0)= -2,
the K of any line perpendicular with the line 2x+y-7=0 is k(1)=1/2
the line perpendicular with 2x+y-7=0 and dawn from (3,4) can be y-4=1/2(x-3)
with the two formulas: 2x+y-7=0
y-4=0.5(x-3)
work out the x,y is the point u want (1.8,3.4)
2007-05-10 20:10:47
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answer #4
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answered by SHUANGYING F 1
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A point on the line is of form (a, 7-2a).
The length from (3,4) to any such point is the sqrt of
(3-a)sq + (4-7+2a)sq =
9 - 6a + a^ + 9 - 12a +4a^ =
18-18a +5a^=
5(a - 1.8)^ + 18 - 5*1.8^;
Minimum length occurs when a = 1.8,
ie at (1.8, 7-3.6=3.4).
2007-05-10 19:27:44
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answer #5
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answered by Sceth 3
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foot? hmmm?
What math level are you?
This can be solved at many levels!
2007-05-10 19:21:40
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answer #6
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answered by Anonymous
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