Question

find the prependicular distance between the lines 3x + 4y-1=0 and 6x+8y+13=0.​

Answer 1

Answer:

3

Step-by-step explanation:

Multiplying eq (1) by 2

we get 6x+8y-2=0

let it be eq(3)

As we can see both the lines are parallel

$|(c1 - c2) \div \sqrt{ {a}^{2} + {b}^{2} } |$

the above formula is for perpendicular distance between two parallel lines

by applying the formula we get distance equal to 15/5=3