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March 25, 2017

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find k if one of the lines given by kx^2+10xy+8y^2=0 is perpendicular to 2x-y=5

  • geometry - ,

    The slope of 2x-y=5 is 2
    So the slope of a perpendicular line is -1/2

    kx^2 + 10xy + 8y^2 = 0 is supposed to be the intersection of 2 straight lines, so let's solve it for y

    y = (-10x ± √(100x^2 - 32kx^2) )/16
    = (-10x ± x√100-68k)/16
    = x(-10 ± √(100-32k) )/16

    So the slope of the two lines is (-10 ± √(100-32k) )/16
    thus:
    (-10 ± √(100-32k) )/16 = -1/2
    -20 ± 2√100-32k = -16
    ±2√100-32k = 4
    ±√100-32k = 2
    square both sides
    100-32k = 4
    32k=96
    k=3

    check my arithmetic.

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