Q.The distance between the lines $y=mx+c_1$ and $y=mx+c_2$ is
(A) $\dfrac{c_1-c_2}{\sqrt{m^2+1}}$
(B) $\dfrac{|c_1-c_2|}{\sqrt{1+m^2}}$
(C) $\dfrac{c_2-c_1}{\sqrt{1+m^2}}$
(D) $0$

The distance between two parallel lines is the perpendicular distance from any point on one line to the other.

Both lines $y = mx + c_1$ and $y = mx + c_2$ have the same slope $m$, so they are parallel. Rewrite them in standard form:

The perpendicular distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by:

Here $A = m$, $B = -1$, $C_1 = c_1$, and $C_2 = c_2$. Substituting: …

The question asks for the distance between two parallel lines. Since both lines have the same slope $m$, they never meet—they're parallel. The "distance" between parallel lines means the perpendicular distance, the shortest gap you'd measure if you dropped a perpendicular from any point on one line to the other.

Why does this approach work? Any two points, one on each line, are separated by some distance, but only the perpendicular distance is constant everywhere along the lines. We can pick a convenient point on one line, find the perpendicular distance to the other, and that's our answer.

Step-by-step derivation

1. Rewrite both lines in standard form.

   The line $y = mx + c_1$ becomes $mx - y + c_1 = 0$, and similarly $y = mx + c_2$ becomes $mx - y + c_2 = 0$.

2. Recall the distance formula from a point to a line.

   The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is

3. Pick a convenient point on the first line.

   Take any point on $mx - y + c_1 = 0$. The easiest choice is $(0, c_1)$, which clearly satisfies $y = m \cdot 0 + c_1$.

4. Find the distance from $(0, c_1)$ to the second line $mx - y + c_2 = 0$.

   Substitute into the distance formula with $A = m$, $B = -1$, $C = c_2$:

5. Simplify the absolute value. …