Question

Question No.5
Values of 'k' so that equation 16^2 + 24xy + ly^2 + kx - 12y - 21 =0, represents a
pair of parallel lines is
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Answer 1

Given: $16x^{2}+24xy+9y^{2}+kx-12y-21=0$

To find: the values of $k$ so thag the given equation represents a pair of straight lines

Rule: before we solve the problem, we must know that the general equation of second degree $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents a pair of straight lines when

$\quad abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$

Solution:

Comparing the given equation with the general equation of second degree, we get

• $a=16,\:b=9,\:h=12,\:g=\frac{k}{2},\:f=-6,\:c=-21$

Putting this values in the condition for pair of straight lines, we get

$\quad 16.9.(-21)+2.(-6).\frac{k}{2}.12-16.(-6)^{2}-9.(\frac{k}{2})^{2}-(-21).12^{2}=0$

$\Rightarrow -3024-72k-576-\frac{9}{4}k^{2}+3024=0$

$\Rightarrow \frac{9}{4}k^{2}+72k+576=0$

$\Rightarrow 9k^{2}+288k+2304=0$

$\Rightarrow k^{2}+32k+256=0$

$\Rightarrow (k+16)^{2}=0$

$\Rightarrow k=-16,\:-16$

Answer: the values of $k$ are $(-16),\:(-16)$

Answer 2

Answer:

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Step-by-step explanation:

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