Straight line , gre prolem $170$ score hard What is the distance between the two straight line represented by the equation

$3x+4y=10$ and $6x+8y=10?$ $$A&gt;1$$ $$B&gt;2$$ $$C&gt;\frac43$$ $$D&gt;\frac12$$ $$E&gt;\frac52$$

I try to solve it , firstly find intercept of both line and then find mid point between of them ,and calculate distance between both line , suggest me where I m wrong
If two lines intersect, then the distance between them is $0$. Otherwise they're parallel and have the same slope, which means the lines can be written as

$ax + by = c$  &nbsp;&nbsp;&nbsp;&nbsp;and
$ax + by = d$

The distance between the lines is then $$\frac{|c-d|}{\sqrt{a^2+b^2}}$$

In your case the two lines are

$3x + 4y = 10$  &nbsp;&nbsp;&nbsp;&nbsp;and
$3x + 4y = 5$

so the distance between them is $\frac{|10-5|}{\sqrt{3^2+4^2}} = 1$.
Since these two lines are parallel lines $( m_{1}=m_{2})$ they would never meet.But their interceptions of Y axis is the constant distance they would maintain.

Just measure that distance $$C_{1}-C_{2} = 2.5-1.25= 1.25 $$
So the ans choice is $A ,&gt;1$.
its not greater than $1.5,1.33,2$ or $2.5$.$(B,C,E)$ though it is greater than $0.5 (1/2)$.