If $k = \frac{a+b}{ab}$ is a non-zero constant,then the point which lies on the straight line $\frac{x}{a} + \frac{y}{b} = 1$ is

Find the equation of a straight line that makes an intercept of $3$ units on the negative direction of the $y$-axis and is inclined at an angle of ${\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$ with the $x$-axis.

$A$ line passing through the point $A(-5, -4)$ intersects the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at points $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,find the equation of the line.

The $x$-intercept and $y$-intercept of the line $2x - 3y = 6$ are respectively:

The equation of the line which makes a right-angled triangle with the axes whose area is $6$ sq. units and whose hypotenuse is $5$ units,is

The equation of the line passing through $(a, b)$ and parallel to the line $\frac{x}{a} + \frac{y}{b} = 1$ is