The point on the line 3x + 4y = 5 which is equidistant from (1, 2) and (3, 4) is

• (7, - 4)

• (15, - 10)

• $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right., \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\right)$

• $\left(0, \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)$

The equanon of the straight line perpendicular to the straight line 3x + 2y = 0 and passing through the point of intersection of the lines x + 3y - 1 = 0 and x - 2y + 4 = 0 is

• 2x - 3y + 1 = 0

• 2x - 3y + 3 = 0

• 2x - 3y + 5 = 0

• 2x - 3y + 7 = 0

The value of $\mathrm{\lambda }$ with $\left|\mathrm{A}\right|$ < 16 such that 2x² - 10xy + 12y² + 5x + $\mathrm{\lambda }$y - 3 = 0 represents a pair of straight lines, is

• - 10

• - 9

• 10

• 9

154.

The area (in square unit) of the triangle formed by x + y + 1 = 0 and the pair of straight lines x² - 3xy + 2y² = 0 is

• $\frac{7}{12}$

• $\frac{5}{12}$

• $\frac{1}{12}$

• $\frac{1}{6}$

The pairs of straight lines x² - 3xy + 2y² = 0 and x - 3xy + 2y² + x - 2 = 0 form a

• Square but not rhombus

• rhombus

• Parallelogram

• Rectangle but not a square

If a straight line L is perpendicular to the line 4x- 2y = 1 and forms a triangle of area 4 sq unit with the coordinate axes, then the equation of the line L is

• 2x + 4y + 7 = 0

• 2x - 4y + 8 = 0

• 2x + 4y + 8 = 0

• 4x - 2y - 8 = 0

$\mathrm{The} \mathrm{image} \mathrm{of} \mathrm{the} \mathrm{point} \left(4, - 13\right) \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{the} \mathrm{line} 5\mathrm{x} + \mathrm{y} + 6 = 0 \mathrm{is}$

• (- 1, - 14)

• (3, 4)

• (1, 2)

• (- 4, 13)

The image of the line x + y - 2 = 0 in the y-axis is

• x - y + 2 = 0

• y - x + 2 = 0

• x + y + 2 = 0

• x + y - 2 = 0

A straight line which makes equal intercepts on positive X and Y axes and which is at a distance 1 unit from the origin intersects the straight line $\mathrm{y} = 2\mathrm{x} + 3 + \sqrt{2} \mathrm{at} \left({\mathrm{x}}_0, {\mathrm{y}}_0\right). \mathrm{Then} 2{\mathrm{x}}_0 + {\mathrm{y}}_0 = ?$

• $3 + \sqrt{2}$

• $\sqrt{2} - 1$

• 1

• 0

The distance between the two lines represented by 8x² - 24xy + 18y² - 6x + 9y - 5 = 0 is

• 0

• $\frac{3}{4\sqrt{13}}$

• $\frac{6}{\sqrt{13}}$

• $\frac{7}{2\sqrt{13}}$