# 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

• The minimum distance? – Jemmy Oct 1 '14 at 21:13
• They do not intersect. The easiest way would be to solve for the y intercept and subtract the two – Eoin Oct 1 '14 at 21:15
• @Eoin: That doesn't work. You need the distance along a perpendicular, and the $y$-axis isn't a perpendicular. – TonyK Jan 1 '17 at 17:26

## 4 Answers

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

$ax + by = c$     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

$3x + 4y = 10$     and
$3x + 4y = 5$

so the distance between them is $\frac{|10-5|}{\sqrt{3^2+4^2}} = 1$.

Hint: They are asking for the shortest distance between the two parallel lines (not the horizontal or vertical distance). Here's an outline for solving it:

• Find the slope (call it $m_1$) of both parallel lines (call them $L_1$ and $L_2$).
• Compute the slope (call it $m_2$) of a line perpendicular to both $L_1$ and $L_2$.
• Choose any point (call it $P$) that is on $L_1$.
• Construct the equation of the line (call it $L$) with slope $m_2$ that passes through $P$.
• Solve a system of equations to find $Q$, the point where $L$ and $L_2$ intersect.
• Use Pythagoras to compute the distance between $P$ and $Q$.

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:

$d_{l1,l2}=\sqrt{(\frac{30}{25}-\frac{15}{25})^2+(\frac{40}{25}-\frac{20}{25})^2}=\sqrt{(\frac{15}{25})^2+(\frac{20}{25})^2}=\sqrt{\frac{625}{625}}=1$

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 $A ,>1$. its not greater than $1.5,1.33,2$ or $2.5$.$(B,C,E)$ though it is greater than $0.5 (1/2)$.