The coordinates of the two points lying on $x+y=4$ and at a unit distance from the straight line $4x+3y=10$ are

Two points $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$. If $P$ is the point of intersection of the lines $4x + 3y + k = 0$ and $3x + 4y + k = 0$,then the range of $b$ is

If $12a + 5b = 9$,where $a, b \in \mathbb{R}$,then the minimum value of $a^2 + b^2$ is:

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$. (in $units$)