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Find the line of the equation that

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 -

    when 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 -

    Hi 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 -

    since 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 -

    I 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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