Question

find the distance between the pair of parallel lines with the given equations.
y = 1/4x + 2
4y - x = -60
a) 17 units
b) √273 or about 16.52 units
c) 4√17 or about 16.49 units
d) 4 units

Explanation:

Step1: Rewrite equations in general form

The first line $y=\frac{1}{4}x + 2$ can be rewritten as $x - 4y+8 = 0$. The second line $4y - x=-60$ can be rewritten as $x - 4y - 60=0$.

Step2: Use distance formula

The distance $d$ between two parallel lines $Ax+By + C_1 = 0$ and $Ax+By + C_2 = 0$ is given by the formula $d=\frac{\vert C_1 - C_2\vert}{\sqrt{A^{2}+B^{2}}}$. Here, $A = 1$, $B=-4$, $C_1 = 8$ and $C_2=-60$.
So, $d=\frac{\vert8-(-60)\vert}{\sqrt{1^{2}+(-4)^{2}}}=\frac{\vert8 + 60\vert}{\sqrt{1 + 16}}=\frac{\vert68\vert}{\sqrt{17}}=\frac{68}{\sqrt{17}}$.

Step3: Simplify the result

Rationalize the denominator: $\frac{68}{\sqrt{17}}=\frac{68\sqrt{17}}{17}=4\sqrt{17}\approx4\times4.123\approx16.49$ units.

Answer:

C) $4\sqrt{17}$ or about 16.49 units