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.
