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Lines x2+4xy+y2=0 and x-y=4 form the figure
- a)a right angle triangle
- b)an isosceles triangle
- c)an equilateral triangle
- d)scalane triangle
Correct answer is option 'C'. Can you explain this answer?
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Lines x2+4xy+y2=0 and x-y=4 form the figurea)a right angle triangleb)a...
Solution:
Given equation of lines are x^2 + 2xy + y^2 = 0, and x - y = 4.
Let's solve the equation of given lines:
x^2 + 2xy + y^2 = 0
Adding x^2 and y^2 on both sides, we get:
x^2 + 2xy + y^2 + x^2 + y^2 = x^2 + y^2
2(x^2 + xy + y^2) = x^2 + y^2
x^2 + y^2 = 2xy
Now, solving this equation with the equation x - y = 4, we get:
x^2 + y^2 = 2xy
(x - y)^2 = 4^2
(x - y) = ±4
Now, we have two cases to consider:
Case 1: x - y = 4
Substituting this value in the equation x^2 + y^2 = 2xy, we get:
x^2 + y^2 = 2xy
(x - y)^2 + 2xy = 2xy
(x - y)^2 = 0
x = y
Hence, the equation of the line becomes x - x = 4, which is not possible.
Therefore, this case is not possible.
Case 2: x - y = -4
Substituting this value in the equation x^2 + y^2 = 2xy, we get:
x^2 + y^2 = 2xy
(x - y)^2 + 2xy = 2xy
(x - y)^2 = 0
x = y
Hence, the equation of the line becomes x - x = -4, which is not possible.
Therefore, this case is also not possible.
Hence, the given lines do not intersect each other.
Therefore, the figure formed by these lines is not a triangle.
Hence, the answer is none of the above options.
Given equation of lines are x^2 + 2xy + y^2 = 0, and x - y = 4.
Let's solve the equation of given lines:
x^2 + 2xy + y^2 = 0
Adding x^2 and y^2 on both sides, we get:
x^2 + 2xy + y^2 + x^2 + y^2 = x^2 + y^2
2(x^2 + xy + y^2) = x^2 + y^2
x^2 + y^2 = 2xy
Now, solving this equation with the equation x - y = 4, we get:
x^2 + y^2 = 2xy
(x - y)^2 = 4^2
(x - y) = ±4
Now, we have two cases to consider:
Case 1: x - y = 4
Substituting this value in the equation x^2 + y^2 = 2xy, we get:
x^2 + y^2 = 2xy
(x - y)^2 + 2xy = 2xy
(x - y)^2 = 0
x = y
Hence, the equation of the line becomes x - x = 4, which is not possible.
Therefore, this case is not possible.
Case 2: x - y = -4
Substituting this value in the equation x^2 + y^2 = 2xy, we get:
x^2 + y^2 = 2xy
(x - y)^2 + 2xy = 2xy
(x - y)^2 = 0
x = y
Hence, the equation of the line becomes x - x = -4, which is not possible.
Therefore, this case is also not possible.
Hence, the given lines do not intersect each other.
Therefore, the figure formed by these lines is not a triangle.
Hence, the answer is none of the above options.
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Lines x2+4xy+y2=0 and x-y=4 form the figurea)a right angle triangleb)an isosceles trianglec)an equilateral triangled)scalane triangleCorrect answer is option 'C'. Can you explain this answer?
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