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Question

If $$3x^{2}+10xy+3y^{2}-15x-21y+k=0$$ represents a pair of straight lines, then the value of $$k$$ is :


A
85
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B
8
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C
18
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D
18
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Solution

The correct option is A $$85$$
$$3{x^2} + 10xy + 3{y^2} - 15x - 21y + k = 0 \to (i)$$
general equation of straight line is
$$a{x^2} + 2fgh - b{g^2} - a{f^2} - c{h^2} = 0 \to $$
$$abc + 2fgh - b{g^2} - a{f^2} - c{h^2} = 0 \to $$
comparing equation 1 with 2, we get
$$a = 3,\,\,h = 5,\,\,b = 3,\,\,g = \dfrac{{ - 15}}{2},f = \dfrac{{ - 21}}{2},c = k$$
putting all values in equation 3, we get
$$\begin{array}{l} 3\times 3\times k+2\times 5\times \left( { \dfrac { { -15 } }{ 2 }  } \right) \times 5-3{ \left( { \dfrac { { -21 } }{ 2 }  } \right) ^{ 2 } }-k{ \left( 5 \right) ^{ 2 } }=0 \\ 9k-375-\dfrac { { 675 } }{ 2 } -\dfrac { { 1323 } }{ 2 } -25k=0 \\ -16k=\dfrac { { 1323 } }{ 2 } +\dfrac { { 675 } }{ 2 } +\dfrac { { 375 } }{ 1 }  \\ -16k=\dfrac { { 2748 } }{ 2 }  \\ k=\dfrac { { +2748 } }{ { 32 } }  \\ k=85.87 \end{array}$$

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