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# 3x2−11xy+10y2−7x+13y+k=0 denotes a pair of straight lines find the value of k

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A)
Solution :
Condition for an equation to represent a pair of straight lines is
$abc+2fgh -af^2-bg^2-ch^2=0$ --------(1)
Given $3x^2-11xy+10y^2-7x+13y+k=0$------(2)
Comparing equation (1) and (2)
$a=3,b=10,c=k$
$h= \large\frac{-11}{2}$
$g= \large\frac{-7}{2}$
$f= \large\frac{13}{2}$
Substituting the values we get,
$3 \times 10 \times k +2 \times \large\frac{13}{2} $$\times \bigg( \large\frac{-7}{2}\bigg) \times \bigg( \large\frac{-11}{2}\bigg)$$-3 \bigg( \large\frac{13}{2}\bigg)^2-10 \bigg( \large\frac{-7}{2}\bigg)^2- k \bigg( \large\frac{-11}{2}\bigg)^2=0$
=> $30 k +\bigg(\large\frac{1001}{4}\bigg) -\frac{507}{4}-\frac{490}{4} -\frac{121 \;k}{4}$
=> $\large\frac{120k +1001-507-490-121\;k}{4}$$=0$
$-k+4=0$

Hence $k=4$