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Hi can you help me in solving this from coordinate geometry?

  1. Mar 5, 2015 #1


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    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    Find the equation of a line passing through point A (1, 2) and whose perpendicular distance from origin is maximum.
    Last edited by a moderator: Mar 5, 2015
  2. jcsd
  3. Mar 5, 2015 #2


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    Start by drawing a sketch showing the coordinate axes and point A.

    Draw an arbitrary line which passes thru point A.

    Can you figure out how to determine the perpendicular distance from this arbitrary line to the origin?
  4. Mar 5, 2015 #3


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    Yes. If that eqn is Ax + By + C = 0
    then the distance from origin is |C|/√(A^2 + B^2)!
  5. Mar 5, 2015 #4


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    Ax + By + C = 0 isn't the whole story.

    A straight line in the plane has two degrees of freedom only, not three like your equation suggests. And this line has to go through point (1,2), so there can't be more than one degree of freedom (e.g. the slope).

    You can eliminate one degree of freedom by requiring A2 + B2 = 1 (effectively dividing by ##\sqrt{A^2+B^2}## -- and no fear of dividing by zero; why not ?)

    That means the new A and B can be written as ##\cos\phi## and ##\sin\phi## for some angle ##\phi## (well, not just 'some' angle...)

    Congrats! you have just derived the so-called 'normal form' equation for a straight line !

    Not there yet: the line has to go through (1,2), so you now eliminate C by substituting the coordinates of point A.

    According to your equation the new C ( let's call it C' ) is then the distance to the origin, so you have that in terms of xA and yA.

    If you know about maximizing a function, you can differentiate that expression and thus find ##\phi##

    If you know about vector products, you can also see that the left hand side is a dot product of two vectors: ##|C'| = (1,2)\;\cdot\; (\cos\phi, \sin\phi)## and the absolute value of that is $$|C'| = |(1,2)| \; |(\cos\phi, \sin\phi)| \; cos\alpha$$ where ##\alpha## is the angle between the two vectors.

    And then |C'| is clearly maximum if ##\alpha = 0## (the two vectors are collinear).
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