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Finding distance of point to y-axis.

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data

    In a cartesian coordinate system, two lines r and s, with angular coefficients 2 and 1/2, respectively, intercept at the origin. If B [itex]\in[/itex] r and C [itex]\in[/itex] s are two points in the first quadrant such that the line segment BC is perpendicular to r and the area of the triangle OBC is 12x10-1, then what is the distance from point B to the y axis?

    2. Relevant equations



    3. The attempt at a solution

    Okay, first I found the relationship between OC and BC using the formula for the area of a triangle:
    [OBC] = 12x10-1 = 12/10 = 6/5 = (OB BC)/2

    Now I am stuck. Whatever I do it seems like I am going around in circles..
     
  2. jcsd
  3. Nov 17, 2012 #2

    Simon Bridge

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    |OB| and |OC| don't have to be the same length.
    But you do know that [OBC] forms a special kind of triangle, and you know one of the apex angles.

    You also need a point Q=(0,y) and look at the triangle [OBQ].
    You know one of these apex angles too.
     
  4. Nov 18, 2012 #3
    Yeah, I know. But the relationship between |BC| and |OB| is that |BC| = 12/(5|OB|); right?

    I did find all the angles of the triangle OBC, but they're not your usual angles and since this problem was taken from a test where calculators are not allowed, you kind of know you're going the wrong way..

    I think I am actually missing some sort of theorem or property that would make this whole problem easier, but I have no idea which.
     
  5. Nov 18, 2012 #4

    Simon Bridge

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    Reformulating - you want the x-coordinate of point B.
     
  6. Nov 18, 2012 #5

    Simon Bridge

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    ... so, rewrite everything in terms of coordinates:

    ##y_r=2x, y_s=\frac{1}{2}x## ... right?
    So the position of the ponts B and C can be writen:
    ##\vec{B}=(x_b, y_b)=(1,2)x_b##
    ##\vec{C}=(x_c,y_c)=(2,1)x_c##

    ... and the area of the triangle [OBC] is:
    ##A=\frac{1}{2}\sqrt{x_b^2+y_b^2}\sqrt{(x_b-x_c)^2+(y_b-y_c)^2}##

    ... that's three equations and four unknowns: you need one more equation: the line through B perpendicular to r, or Pythagoras on the sides of [OBC]?
     
    Last edited: Nov 18, 2012
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