Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$.

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.

The number of lines which pass through the point $(2, -3)$ and are at a distance of $8$ from the point $(-1, 2)$ is:

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:

Statement $(A)$: The points $(2, 1)$ and $(-3, 5)$ lie on opposite sides of the line $3x - 2y + 1 = 0$.
Reason $(R)$: The algebraic perpendicular distances from the given points to the line have opposite signs.

Let $A$ be the point of intersection of the lines $3x + 2y = 14$ and $5x - y = 6$. Let $B$ be the point of intersection of the lines $4x + 3y = 8$ and $6x + y = 5$. The distance of the point $P(5, -2)$ from the line $AB$ is: