(A) The given equation is $9x^{2}-12xy+4y^{2}=0$. Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=9$,$2h=-12$ (so $h=-6$),and $b

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(A) The given equation is $9x^{2}-12xy+4y^{2}=0$. Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=9$,$2h=-12$ (so $h=-6$),and $b=4$. To determine the nature of the lines,we calculate $h^{2}-ab$: $h^{2}-ab = (-6)^{2} - (9 \times 4) = 36 - 36 = 0$. Since $h^{2}-ab=0$,the lines represented by the equation are coincident.

Class 11 Mathematics Pair of straight lines Equation of pair of straight lines

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The straight lines represented by the equation $9x^{2}-12xy+4y^{2}=0$ are

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A
coincident
B
perpendicular
C
intersect at $60^{\circ}$
D
parallel

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The straight lines represented by the equation $9x^{2}-12xy+4y^{2}=0$ are

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