Results 1 to 3 of 3

Math Help - Analytic Geometry Q2

  1. #1
    Member looi76's Avatar
    Joined
    Jan 2008
    Posts
    185

    Analytic Geometry Q2

    Question:
    Show that A(-2,1) , B(5,-2) and C(3,3) are vertices of an equilateral triangle.

    Attempt:

    Distance between points A(-2,1) and B(5,-2):

    AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
    AB = \sqrt{(5-(-2))^2 + (-2-1)^2}
    AB = \sqrt{49 + 9}
    AB = \sqrt{58}

    Distance between points B(5,-2) and C(3,3):

    BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
    BC = \sqrt{(3-5)^2+(3-(-2))^2}
    BC = \sqrt{4 + 25}
    BC = \sqrt{29}

    Distance between points A(-2,1) and C(3,3):

    AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
    AC = \sqrt{(3-(-2))^2 + (3-1)^2}
    AC = \sqrt{25 + 4}
    AC = \sqrt{29}

    I have found out the distance between the points. How can I find out if it is a right angled triangle?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    This is not an equilateral triangle( all the sides are not equal)
    But it's a right triangle because
    AB^2= BC^2+ AC^2 (Pythagoras theorem)

    Pythatgoras Theorem states that the square of the hypotnuse(side opp. to right angle) of a right triangle is equal to the sum of squares of sides
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    Quote Originally Posted by looi76 View Post
    How can I find out if it is a right angled triangle?
    Well, most easily by using Pythagorean theorem, so it is because \overline{AC}^2+\overline{BC}^2=\overline{AB}^2\Le  ftrightarrow 29+29=58.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analytic Geometry Q7
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 31st 2009, 05:25 AM
  2. Analytic Geometry Q6
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 31st 2009, 04:14 AM
  3. Analytic Geometry Q5
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 31st 2009, 12:53 AM
  4. Analytic Geometry Help!
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 8th 2007, 05:13 AM
  5. help on analytic geometry
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 1st 2006, 09:42 AM

Search Tags


/mathhelpforum @mathhelpforum