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Coordinate Geometry II HARD

  1. Mar 13, 2005 #1


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    Coordinate Geometry II HARD!!!

    1a). TriangleABC has vertices A(4,6), B(-3,4) and C(-1,-3). Write down the three inequalities whose intersection is the interior of triangleABC.

    1b). Prove that triangleABC is a right angled isosceles triangle.

    2a). The lines L1 and L2 have the equations 3x-4y+15=0 and 2x+3y-6=0 and intersect at the point P. Write in terms of a constant k, the equation of an arbitrary line through P. (Do not solve the equations simultaneously).

    b). Given that the line through P, L3, also passes through Q(1,1) find the equation of L3.

    i'm not sure if anyone could do it, but im like SOOOO stuck on it and i don't even have a first step..please need help
    Thanks in advance.
  2. jcsd
  3. Mar 14, 2005 #2


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    Well, first draw the three lines, accurately on a graph and label everything. Then consider the equation of the line going from A to C. You can put it in the y=mx+b form:

    [tex]\overline{AC}\rightarrow y_1=m_1x+b_1[/tex]

    Same dif for other two. Just label them with sub-scripts 2 and 3 to keep track of everything.

    Now, consider the first line: Isn't the interior of the triangle contained in the following inequality:

    [tex]y_1\geq m_1x+b_1[/tex]

    You can do the other two right?

    Then the interior of the triangle is just the intersection of the y's right.

    For 1b: An isosceles triangel has two equal sides. Can you not just calculate the length of each line segment and show that two are equal?

    2a) How about using the point-slope form of a line to do that one?

    2b) Use the two-point form of a line.
  4. Mar 14, 2005 #3


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    i understand everything other than question 1a).
    so if you can, would you be able to elaborate please.

    thanks alot salty dog
  5. Mar 15, 2005 #4


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    Aek, just consider one line in the coordinate plane:

    [tex]y=2x+1 [/tex]

    Now consider the inequality:

    [tex]y_1> 2x+1[/tex]

    This inequality represents everything above the line right? That's because every point above the line will have a y component greater than 2x+1.

    Now consider a line above that one:


    But this time consider the inequality:

    [tex]y_2< 2x+2[/tex]

    That inequality represents everything below the second line right?

    Now, consider the intersection of these two sets (everything above the first line, intersection with everything below the second line:
    [tex]y_1\cap y_2[/tex]

    Isn't that everything in-between the two lines?

    Now, you can take the three lines representing the triangle above, form the appropriate inequalities and then just take the intersection of the three to get all the points inside the triangle.

    Hope that helps.
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