Consider three points $$P = ( - \sin (\beta - \alpha ), - cos\beta ),Q = (cos(\beta - \alpha ),\sin \beta )$$ and $$R = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))$$ where $$0 < \alpha ,\beta ,\theta < {\pi \over 4}$$. Then :
Consider the lines given by:
$${L_1}:x + 3y - 5 = 0$$
$${L_2}:3x - ky - 1 = 0$$
$${L_3}:5x + 2y - 12 = 0$$
Match the Statement/Expressions in Column I with the Statements/Expressions in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | L$$_1$$, L$$_2$$, L$$_3$$ are concurrent, if | (P) | $$K = - 9$$ |
| (B) | One of L$$_1$$, L$$_2$$, L$$_3$$ is parallel to atleast one of the other two, if | (Q) | $$K = - {6 \over 5}$$ |
| (C) | L$$_1$$, L$$_2$$, L$$_3$$ form a triangle, if | (R) | $$K = {5 \over 6}$$ |
| (D) | L$$_1$$, L$$_2$$, L$$_3$$ do not form a triangle, if | (S) | $$K = 5$$ |
Let a and b be non-zero real numbers. Then, the equation
$$(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$$ represents :
Let $$\mathrm{O(0,0), P(3,4), Q(6,0)}$$ be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are
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