Answer :
To find the distance between the two given parallel lines, [tex]\(\ell_1\)[/tex] and [tex]\(\ell_2\)[/tex], we can use the formula for the distance between two parallel lines of the form [tex]\( ax + by + c1 = 0 \)[/tex] and [tex]\( ax + by + c2 = 0 \)[/tex]. The formula is:
[tex]\[ \text{Distance} = \frac{|c2 - c1|}{\sqrt{a^2 + b^2}} \][/tex]
First, let's rewrite both equations in the form [tex]\( ax + by + c = 0 \)[/tex]:
1. For line [tex]\(\ell_1\)[/tex] (which is [tex]\( x + y = 0 \)[/tex]):
[tex]\[ x + y + 0 = 0 \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c1 = 0 \)[/tex].
2. For line [tex]\(\ell_2\)[/tex] (which is [tex]\( y = -x - 2 \)[/tex]):
[tex]\[ x + y + 2 = 0 \][/tex]
This can be rearranged to [tex]\( x + y + 2 = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c2 = -2 \)[/tex].
Now substituting into the distance formula:
1. Calculate the absolute difference between [tex]\( c2 \)[/tex] and [tex]\( c1 \)[/tex]:
[tex]\[ |c2 - c1| = |-2 - 0| = 2 \][/tex]
2. Calculate the denominator, which involves [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414 \][/tex]
3. Substitute these values into the formula:
[tex]\[ \text{Distance} = \frac{2}{\sqrt{2}} \approx \frac{2}{1.414} \approx 1.414 \][/tex]
4. Finally, round the distance to the nearest tenth:
[tex]\[ 1.414 \approx 1.4 \][/tex]
So, the distance between the two parallel lines [tex]\(\ell_1\)[/tex] and [tex]\(\ell_2\)[/tex] is approximately 1.4.
Therefore, the answer is:
[tex]\[ \boxed{1.4} \][/tex]
[tex]\[ \text{Distance} = \frac{|c2 - c1|}{\sqrt{a^2 + b^2}} \][/tex]
First, let's rewrite both equations in the form [tex]\( ax + by + c = 0 \)[/tex]:
1. For line [tex]\(\ell_1\)[/tex] (which is [tex]\( x + y = 0 \)[/tex]):
[tex]\[ x + y + 0 = 0 \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c1 = 0 \)[/tex].
2. For line [tex]\(\ell_2\)[/tex] (which is [tex]\( y = -x - 2 \)[/tex]):
[tex]\[ x + y + 2 = 0 \][/tex]
This can be rearranged to [tex]\( x + y + 2 = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c2 = -2 \)[/tex].
Now substituting into the distance formula:
1. Calculate the absolute difference between [tex]\( c2 \)[/tex] and [tex]\( c1 \)[/tex]:
[tex]\[ |c2 - c1| = |-2 - 0| = 2 \][/tex]
2. Calculate the denominator, which involves [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414 \][/tex]
3. Substitute these values into the formula:
[tex]\[ \text{Distance} = \frac{2}{\sqrt{2}} \approx \frac{2}{1.414} \approx 1.414 \][/tex]
4. Finally, round the distance to the nearest tenth:
[tex]\[ 1.414 \approx 1.4 \][/tex]
So, the distance between the two parallel lines [tex]\(\ell_1\)[/tex] and [tex]\(\ell_2\)[/tex] is approximately 1.4.
Therefore, the answer is:
[tex]\[ \boxed{1.4} \][/tex]
