A point equidistant from the lines $4 x+3 y+10=0,5 x-12 y+26=0$ and $7 x+24 y-50=0$ is
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.