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If the three distinct lines x + 2ay + a = 0, x +3by +b = 0 and x + 4ay + a = 0 are concurrent, then the point (a, b) lies on a :
- a)circle
- b)hyperbola
- c)straight line
- d)parabola
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If the three distinct lines x + 2ay + a = 0, x +3by +b = 0 and x + 4ay...
Write the matrix form of the given equations.
By gauss elimination, apply the first row transformation
R2 → R2− R1 and R3 → R3 −R1.
Compute the determinant of the above matrix as,
a = b
The locus of (a, b) lies at x = 0 or y = x. Hence, the point (a, b) will lie on a straight line.
By gauss elimination, apply the first row transformation
R2 → R2− R1 and R3 → R3 −R1.
Compute the determinant of the above matrix as,
a = b
The locus of (a, b) lies at x = 0 or y = x. Hence, the point (a, b) will lie on a straight line.
Most Upvoted Answer
If the three distinct lines x + 2ay + a = 0, x +3by +b = 0 and x + 4ay...
Concurrent Lines:
Three lines are said to be concurrent if they intersect at a common point.
Given Lines:
The given lines are:
1) x - 2ay - a = 0
2) x - 3by - b = 0
3) x - 4ay - a = 0
Concurrent Condition:
For the given lines to be concurrent, the determinant of the coefficients of x, y, and constants in the general equation of lines should be zero.
Determinant:
The determinant of the given lines is:
|1 -2a -a |
|1 -3b -b |
|1 -4a -a |
Calculating the Determinant:
The determinant can be calculated as follows:
|1 -2a -a |
|1 -3b -b |
|1 -4a -a | = (1 * (-3b * (-a)) + (-2a) * (-b) * (-a) + (-a) * (-b) * (-4a)) - ((-a) * (-3b) * (-a) + (-b) * (-4a) * 1 + (-a) * (-2a) * 1)
Simplifying the above expression, we get:
= (3ab - 2a^2 - 4a^2) - (3ab + 4a^2 + 2a^2)
= -6a^2 - 6ab
Condition for Concurrent Lines:
For the given lines to be concurrent, the determinant should be equal to zero.
-6a^2 - 6ab = 0
-6a(a + b) = 0
Conclusion:
From the above equation, we can conclude two cases:
1) -6a = 0, which implies a = 0
2) (a + b) = 0, which implies a = -b
Result:
The point (a, b) lies on a straight line, as the condition for the given lines to be concurrent leads to a linear equation. Hence, the correct answer is option 'C' - straight line.
Three lines are said to be concurrent if they intersect at a common point.
Given Lines:
The given lines are:
1) x - 2ay - a = 0
2) x - 3by - b = 0
3) x - 4ay - a = 0
Concurrent Condition:
For the given lines to be concurrent, the determinant of the coefficients of x, y, and constants in the general equation of lines should be zero.
Determinant:
The determinant of the given lines is:
|1 -2a -a |
|1 -3b -b |
|1 -4a -a |
Calculating the Determinant:
The determinant can be calculated as follows:
|1 -2a -a |
|1 -3b -b |
|1 -4a -a | = (1 * (-3b * (-a)) + (-2a) * (-b) * (-a) + (-a) * (-b) * (-4a)) - ((-a) * (-3b) * (-a) + (-b) * (-4a) * 1 + (-a) * (-2a) * 1)
Simplifying the above expression, we get:
= (3ab - 2a^2 - 4a^2) - (3ab + 4a^2 + 2a^2)
= -6a^2 - 6ab
Condition for Concurrent Lines:
For the given lines to be concurrent, the determinant should be equal to zero.
-6a^2 - 6ab = 0
-6a(a + b) = 0
Conclusion:
From the above equation, we can conclude two cases:
1) -6a = 0, which implies a = 0
2) (a + b) = 0, which implies a = -b
Result:
The point (a, b) lies on a straight line, as the condition for the given lines to be concurrent leads to a linear equation. Hence, the correct answer is option 'C' - straight line.
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If the three distinct lines x + 2ay + a = 0, x +3by +b = 0 and x + 4ay + a = 0 are concurrent, then the point (a, b) lies on a :a)circleb)hyperbolac)straight lined)parabolaCorrect answer is option 'C'. Can you explain this answer?
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