Q111 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
Distance between $8x + 15y - 20 = 0$ and $8x + 15y + 14 = 0$ is:
- A$\dfrac{4}{7}$ units
- Correct answerB$2$ units
- C$\dfrac{10}{17}$ units
- D$34$ units
Solution
The given parallel lines are $8x + 15y - 20 = 0$ and $8x + 15y + 14 = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Substituting $A = 8$, $B = 15$, $C_1 = -20$, and $C_2 = 14$:
$d = \dfrac{|-20 - 14|}{\sqrt{8^2 + 15^2}}$
$d = \dfrac{|-34|}{\sqrt{64 + 225}}$
$d = \dfrac{34}{\sqrt{289}}$
$d = \dfrac{34}{17} = 2$ units.
Answer: $2$ units
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Substituting $A = 8$, $B = 15$, $C_1 = -20$, and $C_2 = 14$:
$d = \dfrac{|-20 - 14|}{\sqrt{8^2 + 15^2}}$
$d = \dfrac{|-34|}{\sqrt{64 + 225}}$
$d = \dfrac{34}{\sqrt{289}}$
$d = \dfrac{34}{17} = 2$ units.
Answer: $2$ units
Hence, the correct answer is option: 'B'.