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1. Find the scalar equation of the line that passes through the point P(0) and has a normal vector n.

P(0)(4,-2), n = (2,7)

2. Find the Cartesian equation of:

(x,y) = (4, -6) + t (8,2)

3. Find the scalr equation of the line that passes through (2,-6) and is parallel to 2x - 3y +8 = 0.

4. Find vector and parametric equations of the line:

5x - 3y + 15 = 0

2007-11-15 12:01:38 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

#1: Since the normal vector is (2, 7), the tangent vector will be (-7, 2), so the slope of the line will be 2/(-7) = -2/7. Therefore using the point-slope equation of the line, we have:

y+2 = -2/7 (x - 4)

Or in standard form:

7y + 14 = -2x + 8
2x + 7y = -6

This could also have been solved by noting that a line of the form ax + by = c will have normal vector (a, b), so we knew this line had to have the form 2x + 7y = c, and then simply used the given point to solve for c.

#2: The slope of the line is 2/8 = 1/4, so using the point-slope form again:

y+6 = 1/4 (x-4)

Or in standard form:

4y+24 = x-4
x - 4y = 28

#3: Since the given line has the form 2x - 3y = c, we know any line parallel to it will have this form as well. Therefore, we need only use the given point to solve for c:

2(2) - 3(-6) = c
4 + 18 = c
c = 22

Thus the equation of the line is 2x - 3y = 22

#4: First we must find a point on the line. Let us find the point at, say, x=0:

-3y + 15 = 0
y=5

So the line passes through (0, 5). Now to find the tangent vector -- we know the _normal_ vector is (5, -3), so rotating it 90° yields a tangent vector of (3, 5). So the vector form of the line is:

(x, y) = (0, 5) + t(3, 5)

And the parametric equations are obtained by just breaking this into its components:

x = 3t
y = 5 + 5t

2007-11-16 06:26:23 · answer #1 · answered by Pascal 7 · 0 0

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2016-10-02 11:09:24 · answer #2 · answered by ? 4 · 0 0

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