Results 1 to 3 of 3

Math Help - equation of lines

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    4

    equation of lines

    Consider the points A(6,-1), B(3,2), C(0,-1).
    (a) Find the equation of the circle S passing through A,B and C (b) Determine the center and radius of S
    (c) Find a fourth point D on S such that ABCD is a square
    (d) Write down the equation of the tangent line to S at D.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Find the point of intersection for the line that is the perpendicular bisector of the line segment AB and the perpendicular bisector of the line segment BC. The point is the center of the circle. The distance from A to that point is the radius of the circle.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    Quote Originally Posted by watcher View Post
    Consider the points A(6,-1), B(3,2), C(0,-1).
    (a) Find the equation of the circle S passing through A,B and C (b) Determine the center and radius of S
    (c) Find a fourth point D on S such that ABCD is a square
    (d) Write down the equation of the tangent line to S at D.
    You can use (c) to "cheat" in (a) and (b)! If (c) is possible, that is, if there exist a point D such that ABCD forms a square, then the two line segments, AB and BC, must be perpendicular and of the same length.

    So check that- The distance from A(6,-1) to B(3,2) is \sqrt{(6-3)^2+ (2+1)^2}= \sqrt{9+ 9}= 3\sqrt{2} and the slope of the line is (2+1)/(3-6)= -1. The distance from B(3,2) to C(0,-1) is \sqrt{(3-0)^2+ (2+1)}= \sqrt{9+ 9}= 3\sqrt{2} and the slope of the line is (2+1)/(3-0)= 1. Yes, A, B, and C do form 3 vertices of a square. But that means that AC is a diagonal of that square and so a diameter of the circle through A, B, and C. The center of the circle is the midpoint of ((6+0)/2, (-1-1)/2)= (3, -1) and the radius is the distance from that point of any of A, B, and C which is also half the distance from A to C, \sqrt{(6-0)^2+ (-1+ 1)^2}/2= 3
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equation of plane from 2 lines
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 3rd 2010, 02:11 AM
  2. Equation of Lines
    Posted in the Geometry Forum
    Replies: 3
    Last Post: February 1st 2009, 09:13 PM
  3. Replies: 2
    Last Post: January 25th 2009, 08:41 PM
  4. Cartesian equation containing the two lines
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 30th 2008, 06:20 AM
  5. Equation of 2 lines
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 16th 2008, 04:35 PM

Search Tags


/mathhelpforum @mathhelpforum