Straight Lines and Pair of Straight Lines

If $(\alpha, \beta)$ is the orthocenter of the triangle $\mathrm{ABC}$ with vertices $A(3,-7), B(-1,2)$ and $C(4,5)$, then $9 \alpha-6 \beta+60$ is equal to :

Let $$(\alpha, \beta)$$ be the centroid of the triangle formed by the lines $$15 x-y=82,6 x-5 y=-4$$ and $$9 x+4 y=17$$. Then $$\alpha+2 \beta$$ and $$2 \alpha-\beta$$ are the roots of the equation :

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 :

A triangle with vertices $$\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right)$$ is :

The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\pi \over 4}$$ at the origin, is equal to :

The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :

The line $$L$$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $$L$$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $$L$$ and $$K$$ is :

The foot of the perpendicular drawn from the origin, on the line, 3x + y = $$\lambda$$ ($$\lambda$$ $$\ne$$ 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :

Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance $${3 \over 5}$$ from the origin. Then which one of the following points lies on any of these lines ?

If p and q are the lengths of the perpendiculars from the origin on the lines, x cosec $$\alpha$$ $$-$$ y sec $$\alpha$$ = k cot 2$$\alpha$$ and x sin$$\alpha$$ + y cos$$\alpha$$ = k sin2$$\alpha$$ respectively, then k² is equal to :