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Posted by on Monday, January 7, 2013 at 8:56am.

A moving point is always equidistant from (5,3) and the line 3x+y+5=0. What is the equation of its locus?

Please help. how to do this problem?

  • geometry - , Monday, January 7, 2013 at 10:02am

    the distance of the point (h,k) from the line ax+by+c=0 is

    d = |ah+bk+c|/√(a^2+b^2)

    the distance between two points (x1, y1) and (x2, y2) is √[ (x2 - x1)^2 + (y2 - y1)^2]

    Let the moving point be (x,y). Substituting the values we have:

    |3x+5y+5|/√10 = √[(x-5)^2 + (y-3)^2]
    (3x+5y+5)^2 = 10((x-5)^2 + (y-3)^2)
    9x^2+30xy+25y^2+30x+50y+25 = 10x^2+10y^2-100x-60y+340

    -x^2 + 30xy + 15y^2 + 130x + 110y - 315 = 0

    Hmm. That's an hyperbola. I was expecting a parabola. Better check my algebra.

  • geometry - , Sunday, January 25, 2015 at 10:04pm

    why did you put square root of 10 in the denmntor? the formula is Ax1+by1+c/+squareroot of A^2+B^2 SUBSTITUTING THE X AND Y IT WILL YOU A REAL NUMBER
    -

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