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Maths - coordinate geometry

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  • bonkysleuth
    bonkysleuth's Avatar
    277 posts since Mar '07
    • 08 May `08, 8:15PM

      the vertice  of triangle ABC are A(01,2), B (1,5) and C(4,3).

       

      a.Find the lengths of the sides AB, BC and CA

      b. what type of triangle is ABC (RIGHT ANGLED ISOCELES)

      c. find the perpendicular distance form B to ac.

      d find the coordinates of the points at which the line AC cuts the x-axis

       

      answer for a.

      AB = square root 13 , same for BC and square root 26 for CA

       

      (C) 2.55 units

      (d) (-11,0)

       

      please help with question d. thanks

  • 16/f/lonely
    16/f/lonely's Avatar
    2,854 posts since Apr '08
    • 08 May `08, 8:51PM

      That's simple. Find the gradient of AC.

      The equation of AC would hence be y = gradient (x) + constant.

      Sub any set of coordinates into the equation and you'll get constant.

      The rest you should know.

  • popikachu
    popikachu's Avatar
    13,163 posts since Dec '06
    • 08 May `08, 9:01PM

      AC is a line...

      A  (1,2)
      C (4,3)

      If this line cut across the x-axis which means y = 0

      First, find the equation of the line...
      y = mx + c

      then sub y in as 0
      Find x.

      Bingo, you have y and you have x now... thats the "coordinate of point" the question is asking ^^

  • popikachu
    popikachu's Avatar
    13,163 posts since Dec '06
    • 08 May `08, 9:06PM

      Answer:

      y = mx +c

      m = (3 - 2) / (4 - 1)
      m = 1/3

      y = (x/3) + c
      2 = 1/3 + c
      c = 5/3

      Equation of line is y = (x/3) + 5/3

      y = 0
      0 = x/3 + 5/3
      0 = x +5
      x = -5

      Coordinate of point (-5, 0)

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