please correct work on finding eqn of perpendicular line

[georgebaseball]

I don't know what I'm doing wrong. The question says "find the equation that goes through (-4,-1) and are perpendicular to the equation through 3x+4y= 6"

So the first think I do is solve for y, to find the slope

. . .3x + 4y = 6
. . .4y = -3x + 6
. . .y = -3/4x + 3/2

So the slope should b 4/3, since it says are perpendicular lines. Next I apply the point-slope equation:

. . .y - y1 = m (x - x1)

. . .y + 1 = (4/3) (x + 4)

Next I multiply by 3 to cancel off the fraction, and I get:

. . .3y + 3 = 4 (x + 4)
. . .3y + 3 = 4x + 16
. . .3y = 4x + 13
. . .y = (4/3)x + 13/3

What am I doing wrong?

 

[georgebaseball]

I know I'm doing something wrong, because I go to this page http://www.algebrahelp.com/calculators/ ... /graphing/ to get a graph of that equation, and i can locate the points (-4,-1)_

 

[stapel]

The question says "find the equation that goes through (-4,-1) and are perpendicular to the equation through 3x+4y= 6"

Wow; that's really badly written. Is the whole book that bad? :shock:

What am I doing wrong?

Your line equation contains the given point, and is perpendicular to the given line. On what basis do you believe your answer to be incorrect?

Thank you.

Eliz.