Question

find the distance between the pair of parallel lines with the given equations.
y = 3x
y = 3x + 10
a) \sqrt{10} or about 3.16 units
b) 10 units
c) 10\sqrt{2} or about 14.14 units
d) 3 units

Explanation:

Step1: Recall the distance formula between two parallel lines

The formula for the distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is \(d=\frac{\vert C_1 - C_2\vert}{\sqrt{A^2 + B^2}}\).

Step2: Rewrite the given lines in standard form

• For \(y = 3x\), rewrite it as \(3x - y + 0 = 0\) (so \(A = 3\), \(B=- 1\), \(C_1 = 0\)).
• For \(y = 3x + 10\), rewrite it as \(3x - y + 10 = 0\) (so \(C_2=- 10\) or we can take \(C_2 = 10\) by rearranging \(3x - y=-10\) to \(3x - y + 10 = 0\); here, \(C_1 = 0\), \(C_2 = 10\)).

Step3: Apply the distance formula

Substitute \(A = 3\), \(B=-1\), \(C_1 = 0\), \(C_2 = 10\) into the formula:
\(d=\frac{\vert 0 - 10\vert}{\sqrt{3^2+(-1)^2}}=\frac{\vert - 10\vert}{\sqrt{9 + 1}}=\frac{10}{\sqrt{10}}=\sqrt{10}\approx3.16\) units.

Answer:

A) $\boldsymbol{\sqrt{10}}$ or about 3.16 units