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The product of perpendiculars from (k, k) to the pair of lines x2 + 4xy + 3y2 = 0 is 4/√5 then k is
- a)±1
- b)±3
- c)±2
- d)±4
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The product of perpendiculars from (k, k) to the pair of lines x2 + 4x...
Correct Answer :- a
Explanation : x2 + 4xy + 3y2 = 0
x2 + xy + 3xy + 3y2 = 0
x(x+y) + 3y(x+y) = 0
(x+3y) (x+y) = 0
Points (k,k)
d1 = |(k+k)/(2)1/2|
2k/(2)1/2
d2 = |(k+3k)/(1+9)1/2|
d2 = |4k/(10)1/2|
d1 * d2 = 4/(5)1/2
|(k+k)/(2)1/2| * |4k/(10)1/2| = 4/(5)1/2
|8k2/(20)1/2| = 4/(5)1/2
8k2/2(5)1/2 = 4/(5)1/2
k2 = +-1
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The product of perpendiculars from (k, k) to the pair of lines x2 + 4x...
To find the product of perpendiculars from (k, k) to the pair of lines, we need to find the equations of the perpendiculars and then find their intersection point.
The given pair of lines is x^2 - 4xy + 3y^2 = 0, which can be factored as (x - y)(x - 3y) = 0. So, the lines are x - y = 0 and x - 3y = 0.
The slope of the line perpendicular to x - y = 0 is the negative reciprocal of the slope of x - y = 0. The slope of x - y = 0 is 1, so the slope of the perpendicular line is -1.
The equation of the line perpendicular to x - y = 0 passing through (k, k) can be written as y - k = -1(x - k).
Simplifying the equation, we get y - k = -x + k.
Rearranging the equation, we get x + y = 2k.
Similarly, the equation of the line perpendicular to x - 3y = 0 passing through (k, k) can be written as y - k = -1(x - k).
Simplifying the equation, we get y - k = -x + k.
Rearranging the equation, we get x + y = 2k.
The two perpendicular lines have the same equation, so they are the same line.
To find the intersection point of the two lines, we can solve the system of equations:
x + y = 2k
x - y = 0
Adding the two equations, we get 2x = 2k.
Dividing by 2, we get x = k.
Substituting x = k into the second equation, we get k - y = 0.
Therefore, the intersection point is (k, k).
The product of the perpendiculars from (k, k) to the pair of lines is 4.
Therefore, the product of perpendiculars from (k, k) to the pair of lines x^2 - 4xy + 3y^2 = 0 is 4.
The given pair of lines is x^2 - 4xy + 3y^2 = 0, which can be factored as (x - y)(x - 3y) = 0. So, the lines are x - y = 0 and x - 3y = 0.
The slope of the line perpendicular to x - y = 0 is the negative reciprocal of the slope of x - y = 0. The slope of x - y = 0 is 1, so the slope of the perpendicular line is -1.
The equation of the line perpendicular to x - y = 0 passing through (k, k) can be written as y - k = -1(x - k).
Simplifying the equation, we get y - k = -x + k.
Rearranging the equation, we get x + y = 2k.
Similarly, the equation of the line perpendicular to x - 3y = 0 passing through (k, k) can be written as y - k = -1(x - k).
Simplifying the equation, we get y - k = -x + k.
Rearranging the equation, we get x + y = 2k.
The two perpendicular lines have the same equation, so they are the same line.
To find the intersection point of the two lines, we can solve the system of equations:
x + y = 2k
x - y = 0
Adding the two equations, we get 2x = 2k.
Dividing by 2, we get x = k.
Substituting x = k into the second equation, we get k - y = 0.
Therefore, the intersection point is (k, k).
The product of the perpendiculars from (k, k) to the pair of lines is 4.
Therefore, the product of perpendiculars from (k, k) to the pair of lines x^2 - 4xy + 3y^2 = 0 is 4.
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The product of perpendiculars from (k, k) to the pair of lines x2 + 4xy + 3y2 = 0 is4/√5 then kisa)±1b)±3c)±2d)±4Correct answer is option 'A'. Can you explain this answer?
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