Let $$C(\alpha, \beta)$$ be the circumcenter of the triangle formed by the lines

$$4 x+3 y=69$$

$$4 y-3 x=17$$, and

$$x+7 y=61$$.

Then $$(\alpha-\beta)^{2}+\alpha+\beta$$ is equal to :

The straight lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$ pass through the origin and trisect the line segment of the line L: $$9 x+5 y=45$$ between the axes. If $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ are the slopes of the lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$, then the point of intersection of the line $$\mathrm{y=\left(m_{1}+m_{2}\right)}x$$ with L lies on :

The combined equation of the two lines $$ax+by+c=0$$ and $$a'x+b'y+c'=0$$ can be written as

$$(ax+by+c)(a'x+b'y+c')=0$$.

The equation of the angle bisectors of the lines represented by the equation $$2x^2+xy-3y^2=0$$ is :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\alpha,\beta)$$, then the quadratic equation whose roots are $$\alpha+4\beta$$ and $$4\alpha+\beta$$, is :