Question

The perpendicular distance between the straight lines
6x+8y+15=0 and 3x+4y+9=0

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Answer 1

Question

The perpendicular distance between the straight lines

6x+8y+15=0 and 3x+4y+9=0

Solution

Given:-

equations of straight lines are ,

• 6x + 8y + 15= 0

Or,

• y = - 3x/4 - 15/8-------(1)

• 3x + 4y + 9 = 0

Or,

• y = -3x/4 - 9/4--------(2)

Find:-

• Distance between these straight line

Explanation

Consider two parallel straight are

• y = mx + c'

and

• y = mx + c"

where,

• m = slope of line

Then, Distance between these line will be (D) = |c" - c' |/√(1+m²)

compare to given equation,

we find here

• c' = -15/8
• c" = -9/4
• m = -3/4

keep all above values

➤▸ D = | (-9/4)-(-15/8)| /√{1+(-3/4)²}]

➤▸ D = | (-18+15)/8 | /√(16+9)/16

➤▸D = |(-3/8) | / √(25/16)

➤▸ D = 3/8 ÷ 5/4

➤▸ D = 3/8 × 4/5

➤▸ D = 3/2 × 5

➤▸ D = 15/2

➤▸ D = 7.5 unit.

Hence:-

• Distance between two straight lines will be = 7.5 units

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