The point (2, 3) is first reflected in the straight line y = x and then translated through a distance of 2 units along the positive direction X-axis. The coordinates of the transformed point are

• (5, 4)

• (2, 3)

• (5, 2)

• (4, 5)

If the straight line 2x + 3y - 1 = 0, x + 2y - 1 = 0 and ax + by - 1 = 0 form a triangle with origin as orthocentre, then (a, b) is equal to

• (6, 4)

• ( - 3, 3)

• ( - 8, 8)

• (0, 7)

The point on the line 4x - y - 2 = 0 which is equidistant from the points ( - 5, 6) and (3, 2) is

• (2, 6)

• (4, 14)

• (1, 2)

• (3, 10)

If the slope of one of the lines represented by ax² - 6xy + y² = 0 is the square of the other,then the value of a is

•  - 27 or 8

•  - 3 or 2

•  - 64 or 27

•  - 4 or 3

The values that m can take, so that the straight line y = 4x + m touches the curve x² + 4y² = 4 is

• $\pm \sqrt{45}$

• $\pm \sqrt{60}$

• $\pm \sqrt{65}$

• $\pm \sqrt{72}$

The angle between the straight lines represented by $\left({\mathrm{x}}^2 + {\mathrm{y}}^2\right){\sin }^2\left(\mathrm{\alpha }\right) = {\left(\mathrm{xcos}\left(\mathrm{\alpha }\right) - \mathrm{ycos}\left(\mathrm{\alpha }\right)\right)}^2$ is

• $\frac{\mathrm{\alpha }}{2}$

• $\mathrm{\alpha }$

• $2\mathrm{\alpha }$

• $\frac{\mathrm{\pi }}{2}$

If the lines x + 3y - 9 = 0, 4x + by - 2 = 0 and 2x - y - 4 = 0 are concurrent, then the equation of the line passing through the point (b, 0) and concurrent with the given lines, is

• 2x + y + 10 = 0

• 4x - 7y + 20 = 0

• x - y + 5 = 0

• x - 4y + 5 = 0

The distance from the origin to the image of (1, 1) with respect to the line x + y + 5 = 0 is

• $7\sqrt{2}$

• $3\sqrt{2}$

• $6\sqrt{2}$

• $4\sqrt{2}$

The equation of the pair of lines joining the origin to the points of intersection of x² + y² = 9 and x + y = 3, is

• ${\mathrm{x}}^2 + {\left(3 - \mathrm{y}\right)}^2 = 9$

• ${\left(3 + \mathrm{y}\right)}^2 + {\mathrm{y}}^2 = 9$

• ${\mathrm{x}}^2 - {\mathrm{y}}^2 = 9$

• xy = 0

Let L be the line joining the origin to the point of intersection of the lines represented by 2x - 3xy - 2y + 10x + 5y = 0. If L is perpendicular to the line kx + y + 3 = 0, then k is equal to

• $\frac{1}{2}$

• $- \frac{1}{2}$

•  - 1

• $\frac{1}{3}$