Question

The distance between parallel lines 3x-4y+7=0 and 6x-8y+10=0 is *
1 point
1/5
2/5
3/5
4/5​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Putting y=0 in 3x−4y+9=0, we get x=−3.

Thus, (−3,0) is a point on the line 3x−4y+9=0.

Length of the perpendicular from (−3,0) to 6x−8y−15=0 is given by

d=

6

2

+(−8)

2

∣−3×6−8×0−15∣

=

10

33