Question

Find the distance between parallel lines 3x+4y =7 and 6x+8y= 10.​

Answer 1

Answer:

Step-by-step explanation:

Put y=0 in 3x+4y−7=0

3x+0−7=0

⇒x=73

∴ Point (73,0) lies on this line

Now, the length of perpendicular from (73,0) to the second line 3x+4y+8=0 is=∣∣∣∣3⋅73+0+832+42‾‾‾‾‾‾‾√∣∣∣∣=∣∣∣155∣∣∣=3 units.

Answer 2

Answer:

distance between two parallel line is 2/5 units

Step-by-step explanation:

Given parallel lines $l_{1}$ =  $3x+4y= 7$  

                                $l_{2}$ =  $6x+8y=10$

$l_{2}$ can be written as  = $3x+4y = 5$    [ divided equation by 2 ]

the parallel lines are  $l_{1} = 3x+4y -7 =0$  

                                   $l_{2} = 3x +4y -5=0$  

Compare both $l_{1}$ and $l_{2}$  with  $ax + by + c_{1} =0$  and $ax + by + c_{2} =0$

⇒  $a =$ 3, $b=$ 4 and $c_{1} =$ -7   $c_{2} =$  -5  

formula for the distance between two parallel lines is given by

d = |$c_{1}$ $-$ $c_{2}$| / $\sqrt{a^{2}+b^{2} }$  

   = |-7 -(-5)| / $\sqrt{3^{2}+4^{2} }$  =   |-7+5| / $\sqrt{9+16}$  =   $\frac{|-2|}{\sqrt{25} }$   = 2/5

the distance between given parallel line is 2/5 units