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Thread: Studying for exam need help!

  1. #1
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    Studying for exam need help!

    Need help with this question from coordinate geometry:

    Write down the equation of any line through the point (-4, 2).
    Hence find the equations of the two lines through the point (-4, 2) whose perpendicular distance from the origin is 2.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by milan View Post
    Need help with this question from coordinate geometry:

    Write down the equation of any line through the point (-4, 2).
    Hence find the equations of the two lines through the point (-4, 2) whose perpendicular distance from the origin is 2.
    Use $\displaystyle y=mx+b$ where b is the y-intercept

    $\displaystyle 2=-4m+b$, now the y-intercept can be 2 or -2 for the two lines...

    $\displaystyle 2=-4m+2$
    $\displaystyle m=0$
    $\displaystyle b=2$

    $\displaystyle 2=-4m-2$
    $\displaystyle m=-1$
    $\displaystyle b=-2$

    You have everything you need!
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  3. #3
    MHF Contributor red_dog's Avatar
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    The equation of the line with the slope $\displaystyle m$ and passing through the point $\displaystyle M_0(x_0,y_0) is$
    $\displaystyle y=y_0=m(x-x_0)$
    In this case $\displaystyle y-2=m(x+4)$ or $\displaystyle mx-y+4m+2=0$

    The distance from the point $\displaystyle M_0(x_0,y_0)$ to the line $\displaystyle ax+by+c=0$ is
    $\displaystyle \displaystyle d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$
    In this case $\displaystyle \displaystyle d=\frac{|4m+2|}{\sqrt{m^2+1}}=2\Rightarrow |2m+1|=\sqrt{m^2+1}$
    Squaring both members and solving the quadratic we get
    $\displaystyle \displaystyle m_1=0, \ m_2=-\frac{4}{3}$
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  4. #4
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    One of our solutions is correct depending on the wording of the question.
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