Question

distance between parallel lines 6x+8y+21=0 and 3x+4y+7=0 is

10/7

-10/7

7/10

-7/10​

Answer 1

we have to find the distance between parallel lines 6x + 8y + 21 = 0 and 3x + 4y + 7 = 0.

solution : concept : when two parallel lines are ax + by + c = 0 and ax + by + k = 0

then, distance between them is given by, |c - k|/√(a² + b²).

here lines are 6x + 8y + 21 = 0 ⇒3x + 4y + 10.5 = 0

3x + 4y + 7 = 0

now distance between them is |10.5 - 7|/√(3² + 4²)

= |3.5|/√(9 + 16)

= 3.5/5

= 0.7

= 7/10

Therefore the distance between given parallel lines is 7/10 unit.

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Answer 2

Answer:

Step-by-step explanation: