Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $l_1$. If a line $l_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $l_1$,then $\left(\frac{k}{h}\right)$ equals

$A$ line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$. Obtain its equation.

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

$A(-2, 3)$ is a point on the line $4x + 3y - 1 = 0$. If the points on the line that are $10$ units away from the point $A$ are $(x_1, y_1)$ and $(x_2, y_2)$,then $(x_1 + y_1)^2 + (x_2 + y_2)^2 =$

If $a, b, c$ are in $A.P.$,then the straight line $ax + by + c = 0$ will always pass through the point

$A$ square of side $a$ lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha, (0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. The equation of its diagonal not passing through the origin is