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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 · answer #1 · answered by Wal C 6 · 0 1

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

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 · answer #3 · answered by Anonymous · 0 0

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 · answer #4 · answered by SHUANGYING F 1 · 0 0

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 · answer #5 · answered by Sceth 3 · 0 0

foot? hmmm?

What math level are you?

This can be solved at many levels!

2007-05-10 19:21:40 · answer #6 · answered by Anonymous · 0 0

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