Line [tex]\( l_1 \)[/tex] has the equation [tex]\( y = -x - 3 \)[/tex] and line [tex]\( l_2 \)[/tex] has the equation [tex]\( y = -x + \frac{1}{2} \)[/tex]. Find the distance between [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex].

Round your answer to the nearest tenth.



Answer :

To find the distance between the two parallel lines [tex]\( y = -x - 3 \)[/tex] and [tex]\( y = -x + \frac{1}{2} \)[/tex], we can use the formula for the distance between two parallel lines given in the standard form [tex]\( ax + by + c = 0 \)[/tex]:

[tex]\[ \text{Distance} = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \][/tex]

First, we rewrite the equations of the lines in the standard form [tex]\( ax + by + c = 0 \)[/tex].

For the line [tex]\( y = -x - 3 \)[/tex]:

[tex]\[ y = -x - 3 \implies x + y + 3 = 0 \][/tex]

Here, the coefficients are:
[tex]\[ a = 1, \quad b = 1, \quad c_1 = 3 \][/tex]

For the line [tex]\( y = -x + \frac{1}{2} \)[/tex]:

[tex]\[ y = -x + \frac{1}{2} \implies x + y - \frac{1}{2} = 0 \][/tex]

Here, the coefficients are:
[tex]\[ a = 1, \quad b = 1, \quad c_2 = -\frac{1}{2} \][/tex]

Next, we plug these coefficients into the distance formula:

[tex]\[ \text{Distance} = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \][/tex]

Substitute the values:

[tex]\[ \text{Distance} = \frac{\left|- \frac{1}{2} - 3\right|}{\sqrt{1^2 + 1^2}} \][/tex]

Simplify the numerator:

[tex]\[ c_2 - c_1 = - \frac{1}{2} - 3 = - \frac{1}{2} - \frac{6}{2} = - \frac{7}{2} \][/tex]

Taking the absolute value:

[tex]\[ |c_2 - c_1| = \left| - \frac{7}{2} \right| = \frac{7}{2} \][/tex]

Now, compute the denominator:

[tex]\[ \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]

So the distance becomes:

[tex]\[ \text{Distance} = \frac{\frac{7}{2}}{\sqrt{2}} = \frac{7}{2} \cdot \frac{1}{\sqrt{2}} = \frac{7}{2} \cdot \frac{\sqrt{2}}{2} = \frac{7\sqrt{2}}{4} \][/tex]

Now, we approximate this value. First, compute [tex]\( \sqrt{2} \)[/tex]:

[tex]\[ \sqrt{2} \approx 1.414 \][/tex]

So:

[tex]\[ \frac{7\sqrt{2}}{4} \approx \frac{7 \cdot 1.414}{4} = \frac{9.898}{4} \approx 2.4748737341529163 \][/tex]

Rounding to the nearest tenth:

[tex]\[ 2.5 \][/tex]

Thus, the distance between the lines [tex]\( \ell_1 \)[/tex] and [tex]\( \ell_2 \)[/tex] is approximately [tex]\( 2.5 \)[/tex] units.