Need some help with this one.

Let L represent the line y = (-4/3)x + 2. Write an equation for the line through p(9,-7) that is parallel to L, and one that is perpendicular to L.

How should I start?

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- Aug 8th 2009, 12:15 PMSåxonEquation for parallel and perpendicular lines
Need some help with this one.

Let L represent the line y = (-4/3)x + 2. Write an equation for the line through p(9,-7) that is parallel to L, and one that is perpendicular to L.

How should I start? - Aug 8th 2009, 12:18 PMeXist
Wrong section first of all. Should be in the pre-calc section I believe, but no worries.

To start, do you know the formula for the equation of a line given the slope of the line and a point on the line?

Also, can you tell me the slope of L given that equation? - Aug 8th 2009, 12:38 PMSåxon
Sorry for the wrong section, I forgot that were still in the review sections of Calculus, but to answer your question, do you mean slope intercept form? y = mx + b? and the slope i think would be -4/3?

- Aug 8th 2009, 12:42 PMeXist
You're correct about the slop being -4/3.

However, if I gave you the point (1,3) and the slope -2, can you form a line with those two things?

Edit: Since I have to go out for a bit, I'll let you know the equation I was looking for is: $\displaystyle (y - y_0) = m(x - x_0)$

Where $\displaystyle (x_0, y_0)$ is a point on the line and m is the slope for that line.

In your problem your looking for a line with the point (9, -7) and a slope that is parallel to $\displaystyle y = (-4/3)x + 2$.

So ask yourself, whats the slope of the line I'm looking for? Then you can use the equation above ($\displaystyle (y - y_0) = m(x - x_0)$). Similar problem for the perpendicular line. Whats the slope of the perpendicular line your looking for? Then you can use the same point and that new slope in the same equation: $\displaystyle (y - y_0) = m(x - x_0)$

Good luck :D be back later

-Chad - Aug 8th 2009, 01:04 PMSåxon
y - (-) 7 = -4/3x - 9?

so y = -4/3x -16, correct?

and the perpendicular slope is just 4/3, so perpendicular would be y = 4/3x -16?

Thanks! - Aug 8th 2009, 03:05 PMeXist
Careful, it should be the quantity $\displaystyle -4/3(x - 9) = -4/3x + 12$ when you distribute through.

So it should be $\displaystyle y + 7 = -4/3x + 12$.

And for the perpendicular, the slope should be the negative reciprocal. So the negative reciprocal of -4/3 = 3/4.