If a pair of lines drawn through the origin forms an isosceles right-angled triangle with the line $2x + 3y = 6$,then those lines are

If the area of the triangle formed by the pair of lines $8x^2-6xy+y^2=0$ and the line $2x+3y=a$ is $7$,then $a$ is equal to

Assertion $(A)$: The lines $2x^2 + 5xy + 2y^2 = 0$ and $x - 2y + 1 = 0$ form a right-angled triangle.
Reason $(R)$: The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of perpendicular lines if $a + b = 0$.
Choose the correct answer.

The product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$ is

If the equation of the pair of lines passing through $(1, 1)$ and perpendicular to the pair of lines $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,then $\frac{b}{a} =$

For an integer $k$,if the area of the triangle formed by the pair of lines $S = 3x^2 - 2kxy + y^2 = 0$ with the line $L = 2x - y - 6 = 0$ is $36$ sq. units,then for the angle $\theta$ between the lines $S = 0$,$\sin \theta =$