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The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:
- a)2x – 3y – 8 = 0
- b)2x + 3y – 5 = 0
- c)4x – 6y + 7 = 0
- d)3x – 2y + 8 = 0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The equation of the line parallel to the line 2x – 3y = 1 and pa...
The midpoint of the line segment is (1+1/2, 3-7/2)
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8
Most Upvoted Answer
The equation of the line parallel to the line 2x – 3y = 1 and pa...
To find the equation of a line parallel to the line 2x - 3y = 1 and passing through the midpoint of the line segment joining the points (1, 3) and (1, 7), we can follow these steps:
Step 1: Determine the slope of the given line
The given line equation is 2x - 3y = 1. We can rewrite this equation in slope-intercept form (y = mx + b) by isolating y:
-3y = -2x + 1
Divide both sides by -3 to get the coefficient of y to be 1:
y = (2/3)x - 1/3
Comparing the equation with y = mx + b, we can see that the slope of the given line is 2/3.
Step 2: Find the midpoint of the line segment
The line segment joining the points (1, 3) and (1, 7) is a vertical line. The x-coordinate remains the same, so the x-coordinate of the midpoint is also 1. To find the y-coordinate of the midpoint, we can average the y-coordinates of the two points:
(3 + 7)/2 = 10/2 = 5
So the midpoint of the line segment is (1, 5).
Step 3: Use the slope and midpoint to find the equation of the parallel line
Since the parallel line has the same slope as the given line, which is 2/3, we can use the point-slope form of a line to find its equation. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we get:
y - 5 = (2/3)(x - 1)
Multiply both sides by 3 to eliminate the fraction:
3y - 15 = 2x - 2
Rearranging the equation:
2x - 3y = 13
Multiplying both sides by -1 to make the coefficient of x positive:
-2x + 3y = -13
Or, rearranging:
2x - 3y + 13 = 0
So, the correct equation of the line parallel to 2x - 3y = 1 and passing through the midpoint of the line segment joining (1, 3) and (1, 7) is 2x - 3y + 13 = 0, which corresponds to option A.
Step 1: Determine the slope of the given line
The given line equation is 2x - 3y = 1. We can rewrite this equation in slope-intercept form (y = mx + b) by isolating y:
-3y = -2x + 1
Divide both sides by -3 to get the coefficient of y to be 1:
y = (2/3)x - 1/3
Comparing the equation with y = mx + b, we can see that the slope of the given line is 2/3.
Step 2: Find the midpoint of the line segment
The line segment joining the points (1, 3) and (1, 7) is a vertical line. The x-coordinate remains the same, so the x-coordinate of the midpoint is also 1. To find the y-coordinate of the midpoint, we can average the y-coordinates of the two points:
(3 + 7)/2 = 10/2 = 5
So the midpoint of the line segment is (1, 5).
Step 3: Use the slope and midpoint to find the equation of the parallel line
Since the parallel line has the same slope as the given line, which is 2/3, we can use the point-slope form of a line to find its equation. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we get:
y - 5 = (2/3)(x - 1)
Multiply both sides by 3 to eliminate the fraction:
3y - 15 = 2x - 2
Rearranging the equation:
2x - 3y = 13
Multiplying both sides by -1 to make the coefficient of x positive:
-2x + 3y = -13
Or, rearranging:
2x - 3y + 13 = 0
So, the correct equation of the line parallel to 2x - 3y = 1 and passing through the midpoint of the line segment joining (1, 3) and (1, 7) is 2x - 3y + 13 = 0, which corresponds to option A.
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The equation of the line parallel to the line 2x – 3y = 1 and pa...
Midpoint of (1,3)and (1,-7)=(1,-2).....Eqn parallel to 2x-3y=1 is 2x-3y=kpassing through (1,-2) is 2(1)-3(-2)=k...=>>k=8...eqn of line parallel to the line 2x-3y=1.....is 2x-3y=8...
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The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer?
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The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer?.
The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:a)2x – 3y – 8 = 0b)2x + 3y – 5 = 0c)4x – 6y + 7 = 0d)3x – 2y + 8 = 0Correct answer is option 'A'. Can you explain this answer?.
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