Question

If $2x^2+7xy+3y^2+8x+14y+k=0$ represents a pair of straight lines, then the value of $k$ is

• A. $5$
• B. $18$
• C. $8$
• D. $16$

Answer 1

The correct option is C $8$

The general equation of second degree is :
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
Given : $2x^2+7xy+3y^2+8x+14y+k=0$
Comparing both,
$a=2,h=\frac{7}{2},b=3,g=4,f=7,c=k$

$\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=6k+2\left(7\right)\left(4\right)\left(\frac{7}{2}\right)-2(7{)}^2-3(4{)}^2-k{\left(\frac{7}{2}\right)}^2$
Now, $\Delta =0$
$\Rightarrow \frac{49k}{4}-6k=196-146=50$
$\Rightarrow k=8$

For $k=8,$
$h^2-ab=\frac{49}{4}-6>0$
hence, represents a pair of straight lines.