Find the equation of the line passing through the intersection of the lines $y = 3$ and $x + y = 0$ and parallel to the line $2x - y = 4$.

If we reduce $3x + 3y + 7 = 0$ to the form $x \cos \alpha + y \sin \alpha = p$,then the value of $p$ is

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

The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

Consider the equation $y-y_1=m(x-x_1)$. If $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$, then

The parallelism condition for two straight lines,one of which is specified by the equation $ax + by + c = 0$ and the other being represented parametrically by $x = \alpha t + \beta$ and $y = \gamma t + \delta$,is given by: