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by **santaclaus** on Thu Mar 12, 2009 1:26 am

Find a formula for the line parallel to the line y=20-4x and containing hte point (3,12)

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by **stapel_eliz** on Thu Mar 12, 2009 2:14 am

santaclaus wrote:Find a formula for the line parallel to the line y=20-4x and containing hte point (3,12)

To learn about the relationships between parallel and perpendicular slopes, try this lesson on slope.

To learn how to use a point and a slope to find the equation for a line, try this lesson on line equations.

by **santaclaus** on Fri Mar 13, 2009 8:49 pm

so, after the lesson I read, could I solve this by saying that -4 is the slope parallel? I don't know how to use the point (3 and 12) to think of an equation y=mx +b? or y-y1 = -4(x-x1) and if I solve, i get y-12 = -4x+12 and then get y= -4x +24????

by **stapel_eliz** on Sat Mar 14, 2009 12:00 am

santaclaus wrote:so, after the lesson I read, could I solve this by saying that -4 is the slope parallel?

Yes; parallel lines have the same slope.

santaclaus wrote:I don't know how to use the point (3 and 12) to think of an equation

Try using the methods explained and demonstrated in the lesson on straight-line equations (in the link provided earlier). :wink:

santaclaus wrote:i get y-12 = -4x+12 and then get y= -4x +24????

I think you did the following:

. . . . . $m\, =\, -4\, \mbox{ and }\, (x_1,\, y_1)\, =\, (3,\, 12)$

. . . . . $y\, -\, y_1\, =\, m(x\, -\, x_1)$

. . . . . $y\, -\, 12\, =\, -4(x\, -\, 3)$

. . . . . $y\, -\, 12\, =\, -4x\, +\, 12$

. . . . . $y\, -\, 12\, +\, 12=\, -4x\, +\, 12\, +\, 12$

If so, your method is fine!

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formula for line parallel to another line

formula for line parallel to another line

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