Q.Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

The distance between two parallel lines can be found by taking any point on one line and computing its perpendicular distance to the other line.

Step 1. Pick a convenient point on the first line $3x - 4y + 7 = 0$. Setting $x = 0$ gives $-4y + 7 = 0$, so $y = \frac{7}{4}$. Thus $(0, \frac{7}{4})$ lies on the first line.

Step 2. Use the point-to-line distance formula. The perpendicular distance from $(x_0, y_0)$ to the line $ax + by + c = 0$ is

$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$ …

Why this approach works

When two lines are parallel, they never meet—they maintain a constant separation everywhere. The distance between them is the length of the perpendicular dropped from any point on one line to the other.

The key insight: since the lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$ have identical coefficients for $x$ and $y$, they are indeed parallel (same normal vector $(3, -4)$). We can use the standard formula for the distance between parallel lines of the form $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$.

This formula comes from taking any point on the first line and computing its perpendicular distance to the second line using the point-to-line distance formula.

Step-by-step solution

1. Verify the lines are parallel

   Both lines have the form $3x - 4y + c = 0$, so their normal vectors are identical: $(3, -4)$. This confirms they are parallel and the distance formula applies.

2. Identify the coefficients

   For the first line $3x - 4y + 7 = 0$: we have $a = 3$, $b = -4$, $c_1 = 7$.

   For the second line $3x - 4y + 5 = 0$: we have $a = 3$, $b = -4$, $c_2 = 5$.

3. Apply the distance formula

   Substitute into the formula: …