Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To find the distance between the two parallel lines, we start by rewriting their equations into a more standard form.
For line [tex]\( \ell_1 \)[/tex]:
[tex]\[ -x + y = 0 \][/tex]
This can be rewritten as:
[tex]\[ y = x \][/tex]
For line [tex]\( \ell_2 \)[/tex]:
[tex]\[ -2x + 2y = 7 \][/tex]
We can divide through by 2 to simplify:
[tex]\[ -x + y = 3.5 \][/tex]
So, this can be rewritten as:
[tex]\[ y = x + 3.5 \][/tex]
Now we have the two lines in slope-intercept form:
[tex]\[ \ell_1: y = x \][/tex]
[tex]\[ \ell_2: y = x + 3.5 \][/tex]
Since these lines are parallel, they have the same slope (m = 1). To find the distance between two parallel lines of the form [tex]\( y = mx + c_1 \)[/tex] and [tex]\( y = mx + c_2 \)[/tex], we can use the formula:
[tex]\[ \text{distance} = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}} \][/tex]
For these lines:
[tex]\[ c_1 = 0 \quad \text{(from } y = x \text{)} \][/tex]
[tex]\[ c_2 = 3.5 \quad \text{(from } y = x + 3.5 \text{)} \][/tex]
[tex]\[ m = 1 \][/tex]
Substitute these values into the distance formula:
[tex]\[ \text{distance} = \frac{|3.5 - 0|}{\sqrt{1 + 1^2}} \][/tex]
Simplify the expression:
[tex]\[ \text{distance} = \frac{3.5}{\sqrt{2}} \][/tex]
We know that [tex]\( \sqrt{2} \approx 1.414 \)[/tex], so:
[tex]\[ \text{distance} \approx \frac{3.5}{1.414} \][/tex]
[tex]\[ \text{distance} \approx 2.4748737341529163 \][/tex]
Rounding to the nearest tenth:
[tex]\[ \text{distance} \approx 2.5 \][/tex]
So, the distance between line [tex]\( \ell_1 \)[/tex] and line [tex]\( \ell_2 \)[/tex] is approximately 2.5 units.
For line [tex]\( \ell_1 \)[/tex]:
[tex]\[ -x + y = 0 \][/tex]
This can be rewritten as:
[tex]\[ y = x \][/tex]
For line [tex]\( \ell_2 \)[/tex]:
[tex]\[ -2x + 2y = 7 \][/tex]
We can divide through by 2 to simplify:
[tex]\[ -x + y = 3.5 \][/tex]
So, this can be rewritten as:
[tex]\[ y = x + 3.5 \][/tex]
Now we have the two lines in slope-intercept form:
[tex]\[ \ell_1: y = x \][/tex]
[tex]\[ \ell_2: y = x + 3.5 \][/tex]
Since these lines are parallel, they have the same slope (m = 1). To find the distance between two parallel lines of the form [tex]\( y = mx + c_1 \)[/tex] and [tex]\( y = mx + c_2 \)[/tex], we can use the formula:
[tex]\[ \text{distance} = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}} \][/tex]
For these lines:
[tex]\[ c_1 = 0 \quad \text{(from } y = x \text{)} \][/tex]
[tex]\[ c_2 = 3.5 \quad \text{(from } y = x + 3.5 \text{)} \][/tex]
[tex]\[ m = 1 \][/tex]
Substitute these values into the distance formula:
[tex]\[ \text{distance} = \frac{|3.5 - 0|}{\sqrt{1 + 1^2}} \][/tex]
Simplify the expression:
[tex]\[ \text{distance} = \frac{3.5}{\sqrt{2}} \][/tex]
We know that [tex]\( \sqrt{2} \approx 1.414 \)[/tex], so:
[tex]\[ \text{distance} \approx \frac{3.5}{1.414} \][/tex]
[tex]\[ \text{distance} \approx 2.4748737341529163 \][/tex]
Rounding to the nearest tenth:
[tex]\[ \text{distance} \approx 2.5 \][/tex]
So, the distance between line [tex]\( \ell_1 \)[/tex] and line [tex]\( \ell_2 \)[/tex] is approximately 2.5 units.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.