Perpendicular equation help

**snickerdoodle27** · Aug 18th 2012, 09:39 AM

Find p such that the lines whose equations are 3px + 8y = 5 and 6y - 4px = -1 are perpendicular.

The answer should be p = 2, -2

But I don't understand how to obtain that.

**adkinsjr** · Aug 18th 2012, 10:35 AM

Right the equations in y=mx+b form. In the first equation you'll have the slope $m_1=\frac{-3p}{8}$ and in the second $m_2=\frac{4p}{6}=\frac{2p}{3}$ . Use what you know about the relationship between the slopes of perpendicular lines to solve for p.

**snickerdoodle27** · Aug 18th 2012, 10:53 AM

Originally Posted by adkinsjr

Right the equations in y=mx+b form. In the first equation you'll have the slope $m_1=\frac{-3p}{8}$ and in the second $m_2=\frac{4p}{6}=\frac{2p}{3}$ . Use what you know about the relationship between the slopes of perpendicular lines to solve for p.

Thank you

For some reason I kept on getting 4, but then I realized I accidentally eliminated the p squared part. Thanks again!!

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