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Thread: equation of st line

  1. #1
    Jun 2009

    equation of st line

    Prove that the equations:
    Q1y^2(cosa+sina)cosa - xy( sin2a - cos2a) +x^2( sina - cosa)sina=0 representsstraight line inclined at 60 to each other .
    Prove also that the area of the triangle formed with them by the straight line (cosa -sina)y -(sina+cosa) x +b=0
    is b^2/(4)
    Q2. Show that the pair
    ax^2 +2hxy +by^2 +( x^2 +y^2) =0 is equally inclined to the same pair.
    Q.3. The vertices of triangle lie on y=xtana,y=xtanb,y=xtanc the circumcenter being origin .prove the locus of intersection of the ortho centre is the line x(sina+sinb+sinc)= y(cosa+cosb+cosc)
    Q4. The base of the triangle passes through (f,g) and the sides are bisected at right angles by x+y=0,y=9x.find the locus of the vertex
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, achiever!

    These are gruesome problems . . . Where did they come from?

    Prove that the equation:

    . . $\displaystyle y^2\cos a(\cos a+\sqrt{3}\sin a) - xy(\sin2a - \sqrt{3}\cos2a) + x^2\sin a(\sin a - \sqrt{3}\cos x) \:=\:0$

    represents straight lines inclined at 60 to each other.
    The equation factors . . . but don't ask me to show the steps.

    $\displaystyle \text{It becomes: }\;\underbrace{\bigg[y\cos a - x\sin a\bigg]}_{L1}\, \underbrace{\bigg[y(\cos a + \sqrt{3}\sin a) - x(\sin a - \sqrt{3}\cos a)\bigg]}_{L_2} \;=\;0 $

    $\displaystyle L_1\!:\;\;y \:=\:\frac{\sin a}{\cos a}x \quad\Rightarrow\quad\boxed{ y \:=\:(\tan a)\,x}$

    $\displaystyle L_2\!:\;\; y \:=\:\frac{\sin a-\sqrt{3}\cos a}{\cos a + \sqrt{3}\sin a}\,x$
    . . Divide top and bottom by $\displaystyle \cos a\!:\;\;y \:=\:\frac{\tan a - \sqrt{3}}{1 + \sqrt{3}\tan a}\,x \:=\:\frac{\tan a - \tan60^o}{1 + \tan60^o\tan a}\,x$
    . . Hence: .$\displaystyle \boxed{y \:=\:\tan(a - 60^o)\,x} $

    Therefore: $\displaystyle L_1$ and $\displaystyle L_2$ form a 60 angle.

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  3. #3
    Jun 2009
    please solve other problems
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