Posted by **B.B.** on Friday, March 26, 2010 at 6:01pm.

Write an equation of the line containing the given point and perpendicular to the given line.

(8,-6); 3x+5y=8. The equation of the line is y=

Here is what I have: 3x+5y=8 5y=-3x+8

y= -3/5(x)+8/5 y=5/3(x)=b -5=5/3(5)+b

-5=25/3+b -15=25+3b

-40=3b b= -40/3= -13 1/3 y= 5/3(x)-13 1/3

Is this correct? If not could you please show me where I went wrong? Thanks.

- Math -
**Reiny**, Friday, March 26, 2010 at 6:06pm
you should know that the slopes of perpendicular lines are opposite reciprocals of each other, so shouldn't the new equation look like

5x - 3y = c ?

plug in the point (8,-6)

40 + 18 = c = 58

so 5x - 3y = 58 , ALL DONE!

(By the way, I did not even make an effort to read your calculations. Since they just run across the page, I can't decide where one statement ends and the next one repeats.)

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