If 3x^2-11xy+10y^2-7x+13y+k=0 denotes a pair | vedclass

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(B) The given equation of the pair of straight lines is $3x^2-11xy+10y^2-7x+13y+k=0$. Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy

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$(3,1)$

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(B) The given equation of the pair of straight lines is $3x^2-11xy+10y^2-7x+13y+k=0$. Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get: $a=3, h=-\frac{11}{2}, b=10, g=-\frac{7}{2}, f=\frac{13}{2}$. The point of intersection $(x, y)$ of the pair of lines is given by the formula: $x = \frac{hf-bg}{ab-h^2}$ and $y = \frac{gh-af}{ab-h^2}$. Calculating the denominator: $ab-h^2 = 3(10) - (-\frac{11}{2})^2 = 30 - \frac{121}{4} = \frac{120-121}{4} = -\frac{1}{4}$. Calculating $x$: $x = \frac{(-\frac{11}{2})(\frac{13}{2}) - (10)(-\frac{7}{2})}{-\frac{1}{4}} = \frac{-\frac{143}{4} + 35}{-\frac{1}{4}} = \frac{\frac{-143+140}{4}}{-\frac{1}{4}} = \frac{-3/4}{-1/4} = 3$. Calculating $y$: $y = \frac{(-\frac{7}{2})(-\frac{11}{2}) - (3)(\frac{13}{2})}{-\frac{1}{4}} = \frac{\frac{77}{4} - \frac{39}{2}}{-\frac{1}{4}} = \frac{\frac{77-78}{4}}{-\frac{1}{4}} = \frac{-1/4}{-1/4} = 1$. Thus,the point of intersection is $(3, 1)$.

If $3x^2-11xy+10y^2-7x+13y+k=0$ denotes a pair of straight lines,then the point of intersection of the lines is

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