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If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation is
- a)x + 5y±5√2 = 0
- b)x - 5 y ± 5 √2 = 0
- c)5x + y ± 5√2 = 0
- d)5 x - y ± 5√2 = 0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If a line is perpendicular to the line 5x - y = 0 and forms a triangle...
5 x - y = 0
y = 5 x ...(1)
Slope = 5
Slope of perpendicular line will be -1/5.
Let equation of line is
......(2)
y = 5 x ...(1)
Slope = 5
Slope of perpendicular line will be -1/5.
Let equation of line is
......(2)Putting y= 0
x = 5 c
OB = 5c
Intersecting point A
x = 5 c
OB = 5c
Intersecting point A
Most Upvoted Answer
If a line is perpendicular to the line 5x - y = 0 and forms a triangle...
To find the equation of the line that is perpendicular to the line 5x - y = 0, we need to find the slope of the given line and then find the negative reciprocal of that slope.
The given line can be rewritten in slope-intercept form as y = 5x.
The slope of this line is 5.
The negative reciprocal of 5 is -1/5.
So, the slope of the perpendicular line is -1/5.
Now, since the line forms a triangle of area 5 square units with the coordinate axes, we can use the formula for the area of a triangle: A = (1/2)bh, where A is the area, b is the base, and h is the height.
In this case, the triangle has a base of 5 units (since the x-intercept of the line is 5) and a height of 5 units (since the y-intercept of the line is 5).
Plugging these values into the area formula, we get:
5 = (1/2)(5)(h)
Simplifying, we get:
5 = (5/2)(h)
Dividing both sides by (5/2), we get:
h = 2
So, the height of the triangle is 2 units.
Now, we can use the point-slope form of a linear equation to find the equation of the perpendicular line. The point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line.
Let's use the point (0, 2) as the point on the line.
Plugging in the values, we get:
y - 2 = (-1/5)(x - 0)
Simplifying, we get:
y - 2 = (-1/5)x
Adding 2 to both sides, we get:
y = (-1/5)x + 2
So, the equation of the line that is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with the coordinate axes is y = (-1/5)x + 2. Therefore, the correct option is a) 5y = -x + 10.
The given line can be rewritten in slope-intercept form as y = 5x.
The slope of this line is 5.
The negative reciprocal of 5 is -1/5.
So, the slope of the perpendicular line is -1/5.
Now, since the line forms a triangle of area 5 square units with the coordinate axes, we can use the formula for the area of a triangle: A = (1/2)bh, where A is the area, b is the base, and h is the height.
In this case, the triangle has a base of 5 units (since the x-intercept of the line is 5) and a height of 5 units (since the y-intercept of the line is 5).
Plugging these values into the area formula, we get:
5 = (1/2)(5)(h)
Simplifying, we get:
5 = (5/2)(h)
Dividing both sides by (5/2), we get:
h = 2
So, the height of the triangle is 2 units.
Now, we can use the point-slope form of a linear equation to find the equation of the perpendicular line. The point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line.
Let's use the point (0, 2) as the point on the line.
Plugging in the values, we get:
y - 2 = (-1/5)(x - 0)
Simplifying, we get:
y - 2 = (-1/5)x
Adding 2 to both sides, we get:
y = (-1/5)x + 2
So, the equation of the line that is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with the coordinate axes is y = (-1/5)x + 2. Therefore, the correct option is a) 5y = -x + 10.
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If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer?
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If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer?.
If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a line is perpendicular to the line 5x - y = 0 and forms a triangle of area 5 square units with co-ordinate axes, then its equation isa)x + 5y±5√2 = 0b)x - 5 y ± 5 √2 = 0c)5x + y ± 5√2 = 0d)5 x - y ± 5√2 = 0Correct answer is option 'A'. Can you explain this answer?.
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