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livestrong136 livestrong136
wrote...
Posts: 324
Rep: 7 0
12 years ago
1. Determine vector and parametric equations for the line through the point A(2, 5) with direction vector  m = (1, 3).

2. Determine parametric equations for the line through (-2, 3) and parallel to the line with vector equation  r = (2, 1) + t(6, 4).

3. Find vector and parametric equations for the line with equation 2x + y + 3 = 0.

4. Determine parametric equations for the line with scalar equation 4x – y + 5 = 0.

5. A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.
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wrote...
12 years ago
5. A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.

If the line is perpendicular, the linear equation will have a slope that is a negative reciprocal of 3x + 2y – 6 = 0.

3x + 2y – 6 = 0

Rearrange:

2y = 6 - 3x

Rearrange to y = mx + b

y = 3 - (3/2)x

The slope of the above equation is -3/2. The slope of the new equation will be 2/3 (the negative reciprocal).

Start creating your new equation using:

y = mx + b

m = 2/3, so:

y = 2/3x + b; now substitute (1, -4), which was given in the question, and solve for "b"

-4 = 2/3(1) + b

-4 - 2/3 = b

-14/3 = b

Therefore, you end up with:

y = 2/3x - 14/3

Hope this helped.
wrote...
12 years ago
Hope this helps, see the attachment Smiling Face with Open Mouth
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livestrong136 Author
wrote...
12 years ago
Thnx Legaspis, can you answer the other questions too. Thnx once again Slight Smile
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