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(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant of t

Vedclass · Accepted Answer

$10/3$

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(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant of the matrix $\begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$.Comparing the given equation $x^2 + kxy + y^2 - 5x - 7y + 6 = 0$ with the general form,we have $a=1, b=1, c=6, h=k/2, g=-5/2, f=-7/2$.Setting the determinant to zero:$\begin{vmatrix} 1 & k/2 & -5/2 \ k/2 & 1 & -7/2 \ -5/2 & -7/2 & 6 \end{vmatrix} = 0$Expanding the determinant:$1(6 - 49/4) - (k/2)(3k - 35/4) - (5/2)(-7k/4 + 5/2) = 0$$(24-49)/4 - (3k^2/2 - 35k/8) + (35k/8 - 25/4) = 0$$-25/4 - 3k^2/2 + 35k/8 + 35k/8 - 25/4 = 0$$-50/4 - 3k^2/2 + 70k/8 = 0$$-25/2 - 3k^2/2 + 35k/4 = 0$Multiplying by $-4$: $6k^2 - 35k + 50 = 0$$(2k - 5)(3k - 10) = 0$Thus,$k = 5/2$ or $k = 10/3$. Given the options,$k = 10/3$ is the correct value.

The equation $x^2 + kxy + y^2 - 5x - 7y + 6 = 0$ represents a pair of straight lines,then $k$ is

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