The distance between the lines 5x - 12y + 65 = 0 and 5x - 12y - 39 = 0 is

Question

The distance between the lines 5x - 12y + 65 = 0 and 5x - 12y - 39 = 0 is (A) 4 (B) 16 (C) 2 (D) 8

Tardigrade Admin · Accepted Answer

The given lines are 5x - 12y + 65 = 0 ...(i) and 5x - 12y - 39 = 0...(ii) Required distance = |(65+39/√25+144)| (∵ d = (|c1-c2|/√a2+b2)) = |(104/13)| = 8 units

Q. The distance between the lines $5 x - 12 y + 65 = 0$ and $5 x - 12 y - 39 = 0$ is

$4$

$16$

$2$

$8$

The given lines are

$5 x - 12 y + 65 = 0... (i)$

and $5 x - 12 y - 39 = 0... (ii)$

Required distance $= ∣ ∣ \frac{65 + 39}{25 + 144} ∣ ∣$

$(∵ d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}})$

$= ∣ ∣ \frac{104}{13} ∣ ∣ = 8$ units

Q. The distance between the lines 5x−12y+65=0 and 5x−12y−39=0 is

4

16

2

8

The given lines are 5x−12y+65=0...(i) and 5x−12y−39=0...(ii) Required distance =∣∣​25+144​65+39​∣∣​ (∵d=a2+b2​∣c1​−c2​∣​) =∣∣​13104​∣∣​=8 units

Q. The distance between the lines $5 x - 12 y + 65 = 0$ and $5 x - 12 y - 39 = 0$ is

$4$

$16$

$2$

$8$

The given lines are

$5 x - 12 y + 65 = 0... (i)$

and $5 x - 12 y - 39 = 0... (ii)$

Required distance $= ∣ ∣ \frac{65 + 39}{25 + 144} ∣ ∣$

$(∵ d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}})$

$= ∣ ∣ \frac{104}{13} ∣ ∣ = 8$ units