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Find an equation of the line that bisects the obtuse angles formed by the lines with equations 3x-y=1 and x+y=-2.
a. (3√2 +√10)x-(√10 + √2)y-2√10+√2=0
b. (3√2 - √10)x+(√10 - √2)y+2√10+√2=0
c. (3√2 + √10)x+(√10 - √2)y+2√10 - √2=0
d. (3√2 + √10)x-(√10 + √2)y-2√10-√2=0
i'm not sure on this one!

  • Math -

    before I answer this question, I have to establish if you know
    1. that slope = tan(angle the line makes with the x-axis)
    2. if m1 and m2 are the slopes of 2 nonparallel lines, and theta is the acute angle between them, then
    tan(theta) = │(m1 - m2)/(1 + m1m2)│

    If not please watch this short video.

    3. tan(2A) = 2tanA/(1-tan^2 A)

    I also noticed that in the first answer given, the slope of the line is
    (3√2+√10)/(√10+√2) which when rationalized is (1+√5)/2 which is the answer that I got for the required line.

    So you might have obtained the correct slope of the new line without realizing it.

    Let me know if you need more help.

  • Math -

    okay thanks

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