The area (in square units) of the triangle formed by the lines 6x2 + 13xy + 6y2 = 0 and x + 2y + 3 = 0 is:
Point (-1, 2) is changed to (a, b) when the origin is shifted to the point (2, -1) by translation of axes. Point (a, b) is changed to (c, d) when the axes are rotated through an angle of 45$^{\circ}$ about the new origin. (c, d) is changed to (e, f) when (c, d) is reflected through y = x. Then (e, f) = ?
If the circle S = 0 cuts the circles x2 + y2 - 2x + 6y = 0, x2 + y2 - 4x - 2y + 6 = 0, and x2 + y2 - 12x + 2y + 3 = 0 orthogonally, then the equation of the tangent at (0, 3) on S = 0 is:
If the area of the circum-circle of the triangle formed by the line 2x + 5y + a = 0 and the positive coordinate axes is \(\frac{29\pi}{4}\) sq. units, then |a| =
The point (a, b) is the foot of the perpendicular drawn from the point (3, 1) to the line x + 3y + 4 = 0. If (p, q) is the image of (a, b) with respect to the line 3x - 4y + 11 = 0, then $\frac{p}{a} + \frac{q}{b} = $