A line passes through $(2,2)$ and is perpendicular to the line $3 x+y=3$. Its $y$-intercept is
The ratio in which the line $3 x+4 y+2=0$ divides the distance between the lines $3 x+4 y+5=0$ and $3 x+4 y-5=0$ is
One vertex of the equilateral triangle with centroid at the origin and one side as $x+y-2=0$ is
If $a, b$ and $c$ are in AP, then the straight lines $a x+b y+c=0$ will always pass through .............. .
Given line is $$a x+b y+c=0\quad \text{.... (i)}$$
Since, $a, b$ and $c$ are in AP, then
$$\begin{aligned} b & =\frac{a+c}{2} \\ a-2 b+c & =0\quad \text{.... (ii)} \end{aligned}$$
On comparing Eqs.(i) and (ii), we get
$$x=1, y=2\quad \text{[using value of b in Eq. (i)]}$$
So, $(1,-2)$ lies on the line.
The line which cuts off equal intercept from the axes and pass through the point $(1,-2)$ is ............ .
Let equation of line is
$$\frac{x}{a}+\frac{y}{a}=1\quad\text{.... (i)}$$
Since, this line passes through $(1,-2)$.
$$\begin{aligned} & \frac{1}{a}-\frac{2}{a}=1 \\ \Rightarrow \quad & 1-2=a \Rightarrow a=-1 \end{aligned}$$
$\therefore$ Required equation of the line is
$$\begin{aligned} -x-y & =1 \\ \Rightarrow \quad x+y+1 & =0 \end{aligned}$$