Question

What is the distance between the lines 3x+4y=9,6x+8y=15 is

Answer 1

distance between two parallel lines ax+by+c1 = 0 & ax+by+c2 = 0
d = |c1-c2|/√(a²+b²)
hence,
d = |9-15/2|/√(3²+4²)
d = |3/2|/5
d = 3/10 
i hope you will understand.
 

Answer 2

Answer:

0.3 units

Step-by-step explanation:

Line 1 : $3x+4y=9$

Slope of line 1= 3

Line 2 : $6x+8y=15$ = $3x+4y=\frac{15}{2}$

Slope of line 2 = 3

Since Slope of both the lines are same .

So, Lines are parallel

Formula of distance between two parallel lines : $d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$

General equation of lines : $A_1x+B_1y= C_1 , A_2x+B_2y=C_2$

So, A = 3

B = 4

$C_1= 9\\C_2=\frac{15}{2}$

$d=\frac{|9-\frac{15}{2}|}{\sqrt{3^2+4^2}}$

$d=0.3$

Hence the distance between the lines 3x+4y=9,6x+8y=15 is 0.3.