# Thread: equation of st line

1. ## 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

2. 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.