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October 1, 2014

October 1, 2014

Posted by **Lynnette** on Thursday, September 23, 2010 at 6:08pm.

hits y=4x^2-20 when x= -5 that is

perpendicular to 5x-6=3y

My solution does not graphical show perpendicular lines, but algebraically it works lol Help!

- PreCalc -
**Reiny**, Thursday, September 23, 2010 at 6:17pmwhen x=-5

y = 4(-5)^2 - 20 = 80

so you have the point (-5,80)

the slope of 5x-6 = 3y is 5/3

so the slope of the required line is -3/5

Now use whichever method you normally use to find the equation of a line given its slope and a point on it.

- PreCalc -
**Lynnette**, Thursday, September 23, 2010 at 6:20pmHi Reiny, Yes, I got that far. I even came up with a line, but am looking to see what others get as my line, when graphed, is not perpendicular to the other line mentioned in the prob (5x-6=3y)

- PreCalc -
**Reiny**, Thursday, September 23, 2010 at 6:29pmsince the new line has slope -3/5, its equation must be

3x + 5y = k

point(-5,80) on it ....

-15 + 400 = k = 385

so new equation is 3x + 5y = 385 or

y = (-3/5)x + 77

Since the slopes are negative reciprocals of each other, their graphs MUST be perpendicular.

Check your calculation of points and check the plotting of points

- PreCalc -
**Lynnette**, Thursday, September 23, 2010 at 6:37pmI hope you're right since that's the equation I got too! I may be entering things into my calculator wrong. Using that is new to me. Thanks for all you're help!

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