Question

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Answer 1

Answer:

Step-by-step explanation:

(iii)

We have,

$x^2+xy-2y^2+4x-y+k=0$

$\implies\,x^2-2y^2+xy+4x-y+k=0$

The given equation is in the form of

$a\,x^2+b\,y^2+2\,hxy+2\,gx+2\,fy+c=0\,\,\,\,\,\,...(1)$

where,

$a=1,\,b=-2,\,h=\dfrac{1}{2},\,g=2,\,f=-\dfrac{1}{2},\,c=k$

Equation (1) will represent a straight line, if,

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

$\implies\left|\begin{array}{ccc}1&\dfrac{1}{2}&2\\\\\dfrac{1}{2}&-2&-\dfrac{1}{2}\\\\2&-\dfrac{1}{2}&k\end{array}\right|=0$

Taking 1/2 common form each row,

$\implies\dfrac{1}{8}\left|\begin{array}{ccc}2&1&4\\1&-4&-1\\4&-1&2k\end{array}\right|=0$

$\implies\left|\begin{array}{ccc}2&1&4\\1&-4&-1\\4&-1&2k\end{array}\right|=0$

$\implies\,2\left|\begin{array}{cc}-4&-1\\-1&2k\end{array}\right|-1\left|\begin{array}{cc}1&-1\\4&2k\end{array}\right|+4\left|\begin{array}{cc}1&-4\\4&-1\end{array}\right|=0$

$\implies\,2(-8k-1)-1(2k+4)+4(-1+16)=0$

$\implies\,-16k-2-2k-4-4+64=0$

$\implies\,-18k+54=0$

$\implies\,18k=54\\\implies\,k=3$