If ax^2+6xy+by^2-10x+10y-6=0 represents a pai | vedclass

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(B) The given equation of the pair of lines is $ax^2+6xy+by^2-10x+10y-6=0$. Comparing this with the general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we g

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$4$

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(B) The given equation of the pair of lines is $ax^2+6xy+by^2-10x+10y-6=0$. Comparing this with the general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=a, H=3, B=b, G=-5, F=5, C=-6$. For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$,which implies $b=-a$. Also,the condition for the general second-degree equation to represent a pair of lines is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Substituting the values: $a(b)(-6) + 2(5)(3)(-5) - a(5)^2 - b(-5)^2 - (-6)(3)^2 = 0$. $-6ab - 150 - 25a - 25b + 54 = 0$. $-6ab - 25(a+b) - 96 = 0$. Since $a+b=0$,we have $-6a(-a) - 25(0) - 96 = 0$. $6a^2 = 96 \Rightarrow a^2 = 16$. Therefore,$|a| = \sqrt{16} = 4$.

If $ax^2+6xy+by^2-10x+10y-6=0$ represents a pair of perpendicular lines,then the value of $|a|$ equals

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