To find the distance between two parallel lines, we start by rewriting their equations in the slope-intercept form [tex]\( y = mx + b \)[/tex].
1. Rewrite the equation of [tex]\( \ell_1 \)[/tex]:
Given: [tex]\( 3x + 2y = 2 \)[/tex]
[tex]\[
\text{Isolate } y: \, 2y = -3x + 2 \implies y = -\frac{3}{2}x + 1
\][/tex]
Thus, [tex]\( \ell_1 \)[/tex] has a slope [tex]\( m_1 = -\frac{3}{2} \)[/tex] and an intercept [tex]\( b_1 = 1 \)[/tex].
2. Analyze the equation of [tex]\( \ell_2 \)[/tex]:
Given: [tex]\( y = -\frac{3}{2}x + \frac{18}{5} \)[/tex]
The slope is [tex]\( m_2 = -\frac{3}{2} \)[/tex] and the intercept is [tex]\( b_2 = \frac{18}{5} \)[/tex].
Since the slopes [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] of the lines are equal, the lines are parallel. We now proceed to determine the distance between these parallel lines. The formula for the distance between two parallel lines [tex]\( y = mx + b_1 \)[/tex] and [tex]\( y = mx + b_2 \)[/tex] is given by:
[tex]\[
\text{Distance} = \frac{|b_2 - b_1|}{\sqrt{1 + m^2}}
\][/tex]
3. Substitute the values into the formula:
[tex]\[
\text{Slope } m = -\frac{3}{2}, \quad b_1 = 1, \quad b_2 = \frac{18}{5}
\][/tex]
Calculate [tex]\( b_2 - b_1 \)[/tex]:
[tex]\[
b_2 - b_1 = \frac{18}{5} - 1 = \frac{18}{5} - \frac{5}{5} = \frac{13}{5}
\][/tex]
Substitute into the distance formula:
[tex]\[
\text{Distance} = \frac{\left|\frac{13}{5}\right|}{\sqrt{1 + \left(-\frac{3}{2}\right)^2}} = \frac{\frac{13}{5}}{\sqrt{1 + \frac{9}{4}}} = \frac{\frac{13}{5}}{\sqrt{\frac{13}{4}}}
\][/tex]
4. Simplify the expression:
[tex]\[
\text{Distance} = \frac{\frac{13}{5}}{\frac{\sqrt{13}}{2}} = \frac{13}{5} \times \frac{2}{\sqrt{13}} = \frac{26}{5 \sqrt{13}} = \frac{26 \sqrt{13}}{65} = \frac{2 \sqrt{13}}{5}
\][/tex]
5. Calculate the numerical value:
[tex]\[
\frac{2 \sqrt{13}}{5} \approx \frac{2 \cdot 3.60555}{5} \approx \frac{7.2111}{5} \approx 1.4422
\][/tex]
6. Round to the nearest tenth:
[tex]\[
1.4422 \approx 1.4
\][/tex]
Therefore, the distance between the lines [tex]\( \ell_1 \)[/tex] and [tex]\( \ell_2 \)[/tex] is approximately [tex]\( 1.4 \)[/tex] units, rounded to the nearest tenth.
