Straight line , gre prolem $170$ score hard What is the distance between the two straight line represented by the equation
$3x+4y=10$ and $6x+8y=10?$ $$A>1$$ $$B>2$$ $$C>\frac43$$ $$D>\frac12$$ $$E>\frac52$$
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 

Let me call the lines $l1$ and $l2$, in which:
$l1: y=\frac{-3}{4}x+2.5$ 
and
$l2: y=\frac{-3}{4}x+1.25$.
With a little notice to the above lines we see that these are the lines mentioned in the question. In another language, we have made the above line equations by altering the shape of the line equations. As we see, both lines have the same slopes, and hence they are parallel to each other and there is no intersection. So we can define distance between them. Let me define $l3$ as:
$l3: y=\frac{4}{3}x$
As we see $l3$ is perpendicular to $l1$ and $l2$ (because the product of their slopes is equal to $-1$).  We wanna find the intersection of $l3$ with the both lines. So we have:
$y=\frac{4}{3}x\rightarrow\frac{4}{3}x=\frac{-3}{4}x+2.5\rightarrow x=\frac{30}{25}\rightarrow y=\frac{40}{25}\rightarrow P1=(\frac{30}{25},\frac{40}{25})$
$y=\frac{4}{3}x\rightarrow\frac{4}{3}x=\frac{-3}{4}x+1.25\rightarrow x=\frac{15}{25}\rightarrow y=\frac{20}{25}\rightarrow P2=(\frac{15}{25},\frac{20}{25})$
In which $P1$ and $P2$ are the intersection of $l3$ with $l1$ and $l2$ respectively. The distance between $P1$ and $P2$ is the distance between $l1$ and $l2$, so we have:
$d_{l1,l2}=\sqrt{(\frac{30}{25}-\frac{15}{25})^2+(\frac{40}{25}-\frac{20}{25})^2}=\sqrt{(\frac{15}{25})^2+(\frac{20}{25})^2}=\sqrt{\frac{625}{625}}=1 $
