Answer :
To find the distance between two parallel lines, we can use the formula for the distance between two lines in the form [tex]\(Ax + By + C = 0\)[/tex]. The formula to find the distance [tex]\(d\)[/tex] between the lines [tex]\(A x + B y + C_1 = 0\)[/tex] and [tex]\(A x + B y + C_2 = 0\)[/tex] is given by:
[tex]\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \][/tex]
Given the equations of the lines:
1. [tex]\(l_1: -x + 2y = 8\)[/tex]
2. [tex]\(l_2: y = \frac{1}{2} x + 1\)[/tex]
We need to rewrite both equations in the standard form [tex]\(Ax + By + C = 0\)[/tex]:
For line [tex]\(l_1\)[/tex]:
[tex]\[ -x + 2y = 8 \quad \Rightarrow \quad -x + 2y - 8 = 0 \][/tex]
Here, [tex]\(A_1 = -1\)[/tex], [tex]\(B_1 = 2\)[/tex], and [tex]\(C_1 = -8\)[/tex].
For line [tex]\(l_2\)[/tex]:
[tex]\[ y = \frac{1}{2} x + 1 \quad \Rightarrow \quad -\frac{1}{2} x + y - 1 = 0 \][/tex]
Multiplying through by 2 to make the coefficients consistent with [tex]\(l_1\)[/tex]:
[tex]\[ -x + 2y - 2 = 0 \][/tex]
Here, [tex]\(A_2 = -1\)[/tex], [tex]\(B_2 = 2\)[/tex], and [tex]\(C_2 = -2\)[/tex].
Now we apply the distance formula:
1. Calculate the numerator [tex]\(|C_2 - C_1|\)[/tex]:
[tex]\[ |C_2 - C_1| = |-2 - (-8)| = |-2 + 8| = 6 \][/tex]
2. Calculate the denominator [tex]\(\sqrt{A^2 + B^2}\)[/tex]:
[tex]\[ \sqrt{A^2 + B^2} = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236 \][/tex]
3. Compute the distance:
[tex]\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} = \frac{6}{\sqrt{5}} \approx \frac{6}{2.236} \approx 2.683 \][/tex]
Rounding the answer to the nearest tenth:
[tex]\[ d \approx 2.7 \][/tex]
Thus, the distance between lines [tex]\(l_1\)[/tex] and [tex]\(l_2\)[/tex] is approximately [tex]\(2.7\)[/tex].
[tex]\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \][/tex]
Given the equations of the lines:
1. [tex]\(l_1: -x + 2y = 8\)[/tex]
2. [tex]\(l_2: y = \frac{1}{2} x + 1\)[/tex]
We need to rewrite both equations in the standard form [tex]\(Ax + By + C = 0\)[/tex]:
For line [tex]\(l_1\)[/tex]:
[tex]\[ -x + 2y = 8 \quad \Rightarrow \quad -x + 2y - 8 = 0 \][/tex]
Here, [tex]\(A_1 = -1\)[/tex], [tex]\(B_1 = 2\)[/tex], and [tex]\(C_1 = -8\)[/tex].
For line [tex]\(l_2\)[/tex]:
[tex]\[ y = \frac{1}{2} x + 1 \quad \Rightarrow \quad -\frac{1}{2} x + y - 1 = 0 \][/tex]
Multiplying through by 2 to make the coefficients consistent with [tex]\(l_1\)[/tex]:
[tex]\[ -x + 2y - 2 = 0 \][/tex]
Here, [tex]\(A_2 = -1\)[/tex], [tex]\(B_2 = 2\)[/tex], and [tex]\(C_2 = -2\)[/tex].
Now we apply the distance formula:
1. Calculate the numerator [tex]\(|C_2 - C_1|\)[/tex]:
[tex]\[ |C_2 - C_1| = |-2 - (-8)| = |-2 + 8| = 6 \][/tex]
2. Calculate the denominator [tex]\(\sqrt{A^2 + B^2}\)[/tex]:
[tex]\[ \sqrt{A^2 + B^2} = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236 \][/tex]
3. Compute the distance:
[tex]\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} = \frac{6}{\sqrt{5}} \approx \frac{6}{2.236} \approx 2.683 \][/tex]
Rounding the answer to the nearest tenth:
[tex]\[ d \approx 2.7 \][/tex]
Thus, the distance between lines [tex]\(l_1\)[/tex] and [tex]\(l_2\)[/tex] is approximately [tex]\(2.7\)[/tex].
