Question

solve
Find the value of k?​

Answer 1

Question :

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

Theory :

An equation of the form $ax{}^{2}+2hxy+by{}^{2}+2gx+2fy+c=0$

is called genral equation of second degree in x and y .

•The necessary and sufficient condition to represent a pair of straight lines is:

Δ =$\left|\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right|$=0

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

Solution ;

Given : Curve equation= $x{}^{2}+kxy+y{}^{2}-5x-7y+6=0$

On camparing with Genral equation:

a=1 , b=1 ,h = k/2 , g =-5/2 ,f =-7/2 and c = 6

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Now it's is given that curve represents a pair of staright lines .

Therefore ; Δ = 0

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

Now put the values :

$\implies6 + 2 \times\frac{ - 7}{2} \times \frac{ - 5}{2} \times \frac{k}{2} - 1 \times (\frac{ - 7}{2}) {}^{2} - 1 \times ( \frac{ - 5}{2} ) { }^{2} - 6 \times ( \frac{k}{2}) {}^{2} = 0$

$\implies6 + \frac{35k}{4} - \frac{49}{4} - \frac{29}{4} - \frac{6k {}^{2} }{4} = 0$

$\implies35k - 50 - 6k {}^{2} = 0$

$\implies6k {}^{2} + 50 - 35k = 0$

$\implies(2k - 5)(3k - 10) = 0$

$\implies \: k = \dfrac{5}{2} \: or \: k = \dfrac{10}{3}$

Correct option b)

${\purple{\boxed{\large{\bold{k=\dfrac{10}{3}}}}}}$