Question

Find the perpendicular distance between the lines
3x+4y-5=0 and 6x+8y-45=0.​

Answer 1

Answer:

$\sf d= 3.5 \ units$

Step-by-step explanation:

Given :

3 x + 4 y - 5 = 0

On comparing with standard equation :

Ax + By + C = 0

Slope of line m = - A / B

m₁ = - 3 / 4

6 x + 8 y - 45 = 0

m₂ = - 6 / 8

m₂ = - 3 / 4

Since slope of both lines are equal.

Therefore they are parallel line.

Now :

Distance between two parallel line :

$\sf d=\left| \dfrac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$

Putting values here :

$\sf d=\left| \dfrac{-45/2+5}{ \sqrt{3^2+4^2}}\right|$

$\sf d=\left| \dfrac{-17.5}{5} \right|$

$\sf d=3.5 \ units$

Therefore we get required answer.

Answer 2

Given : parallel lines 3x + 4y = 5 and   6x + 8y = 45  

To Find  :  Distance between parallel lines

Solution :

3x + 4y = 5  

6x + 8y = 45

Slope of the line =  - 3/4

Hence Slope of perpendicular line =  4/3

Let say a line  y = 4x/3    cuts both the line

3x + 4y = 5  

=> 3x + 4(4x/3) = 5

=> 9x + 16x  = 15

=> 25x  = 15

=> x  = 3/5

    y = 4/5

6x + 8y = 45

=> 6x +8(4x/3) = 45

=> 18x + 32x  = 135

=> 50x  = 135

=> 10x = 27

=> x  = 27/10

    y =  36/10

Points are ( 3/5 , 4/5)  and ( 27/10 , 36/10)

Distance = √(27/10 - 3/5)²  + (36/10 - 4/5)²

= (1/10)√ 21² + 28²

= (7/10)√3² + 4²

= (7/10)5

= 35/10

= 3.5

perpendicular distance between the lines 3x+4y-5=0 and 6x+8y-45=0​ = 3.5

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