Question

The perpendicular distance between the lines $3x+4y-6=0$ and $6x+8y+18=0$ is

• A. 15
• B. 12
• C. 9
• D. 3
• E. 0

Hint:

Logic Tip: Always ensure the $x$ and $y$ coefficients are exactly the same before applying the parallel lines distance formula. If you used $C_1 = -6$ and $C_2 = 18$ without adjusting, you would get an incorrect answer.

Answer 1

The Correct Option is D

Concept:
The perpendicular distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula: $$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$ Before using the formula, the coefficients of $x$ and $y$ must be made identical for both lines.
Step 1: Standardize the equations of the lines.
Line 1: $3x + 4y - 6 = 0$ Line 2: $6x + 8y + 18 = 0$ Divide Line 2 by 2 to match the coefficients of Line 1: $$3x + 4y + 9 = 0$$
Step 2: Identify the components for the distance formula.
From the standardized equations: $A = 3$ $B = 4$ $C_1 = -6$ $C_2 = 9$
Step 3: Calculate the distance.
Substitute the values into the distance formula: $$d = \frac{|-6 - 9|}{\sqrt{3^2 + 4^2}}$$ $$d = \frac{|-15|}{\sqrt{9 + 16}}$$ $$d = \frac{15}{\sqrt{25}}$$ $$d = \frac{15}{5} = 3$$