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So there is: $$\begin{align} e: 5x-2y &= 10 \\ f: 5x-2y &= -19 \end{align} $$ where $e$ and $f$ are parallel lines.

Question: What is the distance of the two parallel lines?

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Have you tried something? –  Patrick Da Silva Apr 9 '12 at 18:13

4 Answers 4

up vote 3 down vote accepted

It is well known that the distance from the point $(x_0, y_0)$ to the line $Ax + By +C = 0$ is $$\dfrac {Ax_0 + By_0 + C}{\sqrt {A^2 + B^2}}$$ The point $(2,0)$ is on line $e$, and the equation for line $f$ is $5x-2y + 19 = 0$. Thus, applying this formula, we obtain $$\dfrac {5(2) + (-2)(0) + 19}{\sqrt {5^2 + 2^2}}$$ $$=\dfrac {29}{\sqrt {29}}$$ $$=\boxed {\sqrt {29}}$$

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Draw the two lines. Let $A=(0,-5)$ be the $y$-intercept of $e$, and let $B=(0,19/2)$ be the $y$-intercept of $f$. Drop a perpendicular from $A$ to the point $C$ on line $f$. The length of the segment $AC$ is what you're looking for.

Now construct a triangle with one vertex at $B$, another vertex at $D=(0, 19/2+5)$, and the last at $E=(2,19/2+5)$. Since the slope of line $f$ is $5/2$, it follows that $E$ lies on line $f$. Also, note $\triangle BDE$ is a right triangle, so the length of the segment $BE$ is $\sqrt{5^2+2^2}=\sqrt{29}$.

Now note that $\triangle ACB$ and $\triangle BDE$ are similar; whence $$ {AC\over AB}={DE\over BE}. $$ But, $AB=5+{19\over2}={29\over2}$, $DE=2$, and $BE=\sqrt{29}$; thus, $$ AC={2\over\sqrt{29}}\cdot{29\over2}=\sqrt{29}. $$

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Here is a yet another proof. Take a point on either line and minimize the distance between them. Choose $x_1$ to be the x coordinate of the point on the first line and $x_2$ the x coordinate of the point on the second line. Then the coordinates of the actual points are $(x_1, \frac{5}{2} x_1 -5)$ and $(x_2, \frac{5}{2} x_2 -\frac{19}{2})$. The distance between these two points is given by:

$d^2 = (x_1-x_2)^2 + (\frac{5}{2}(x_1-x_2) - \frac{29}{2})^2$

Now choose $(x_1, x_2)$ to minimize this distance. Since only terms of the form $(x_1-x_2)$ appear, we can just use $x=(x_1-x_2)$ to simplify things a little. Then we have $d^2 = x^2 + (\frac{5}{2}x - \frac{29}{2})^2$. Minimizing the distance $d$ is the same as minimizing $d^2$, so we can just find out where $d^2$ is a minimum. Differentiating the expression with respect to x and setting it to zero yields $\frac{\partial (d^2)}{\partial x} = \frac{29}{2} x -\frac{145}{2} $. From this we have the minimizing $x= \frac{145}{29}$. Substituting this into the formula gives $d^2 = 29$ from which the result follows.

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$\sqrt{29}$.

The normal to the lines is $N:=\left(\array{5 \\-2}\right)$ which has length $\sqrt{5^2+2^2}=\sqrt{29}$.

(If a line is given by $<n,x>= d$, $<,>$ denoting the scalar product, and $n$ is the unit normal, the signed distance from the line to the origin is $d$. In your case $n=\frac{1}{\sqrt{29}}N$ and the lines are defined by $$ <n,x> = \frac{10}{\sqrt{29}} $$ and $$ <n,x> = -\frac{19}{\sqrt{29}}$$

Because of the different signs they are on 'opposite' sides of the origin, so the distance between the lines is $$\frac{10}{\sqrt{29}}+ \frac{19}{\sqrt{29}}$$:-) ).

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