Question

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

• A. $\frac{5}{3}$
• B. $\frac{10}{3}$
  
• C. $\frac{3}{2}$
• D. $\frac{3}{10}$
  

Answer 1

Answer: (B)

$\frac{10}{3}$

Given: Equationof the pair of straight lines is

$x^2+kxy+y^2-5x-7y+6=0$

We know that the standard equation of the pair of straight lines is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0.$

Comparing the given equation with the standard equation, we get

$a=1,b=1,c=6,f=-\frac{7}{2},g=-\frac{5}{2}andh=\frac{k}{2}$

We also know that the condition for the pair of straight lines is

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

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

$\Rightarrow 6+(-7)(-\frac{5}{2})\times \left(\frac{k}{2}\right)-\frac{49}{4}-\frac{25}{4}-\frac{{3k}^2}{2}=0$

$\Rightarrow -\frac{{3k}^2}{2}+\frac{35k}{4}-\frac{25}{2}=0\Rightarrow 6k^2-35k+50=0\Rightarrow (2k-5)(3k-10)=0\Rightarrow k=\frac{10}{3}$