If the equation of the pair of straight lines | vedclass

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(C) The given pair of lines is $3x^2+11xy-4y^2=0$. Factoring this,we get $(3x-y)(x+4y)=0$. The slopes of these lines are $m_1=3$ and $m_2=-\frac{1}{4}

Vedclass · Accepted Answer

$-19$

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(C) The given pair of lines is $3x^2+11xy-4y^2=0$. Factoring this,we get $(3x-y)(x+4y)=0$. The slopes of these lines are $m_1=3$ and $m_2=-\frac{1}{4}$. The lines perpendicular to these will have slopes $m_1'=-\frac{1}{3}$ and $m_2'=-4$. Since these lines pass through $(1,1)$,their equations are: $y-1=-\frac{1}{3}(x-1) \Rightarrow x+3y-4=0$ $y-1=-4(x-1) \Rightarrow 4x-y-3=0$ The combined equation is $(x+3y-4)(4x-y-3)=0$. Expanding this: $4x^2-xy-3x+12xy-3y^2-9y-16x+4y+12=0$. Simplifying: $4x^2+11xy-3y^2-19x-5y+12=0$. Comparing with $ax^2+2hxy+by^2+2gx+2fy+12=0$,we get $a=4, 2h=11, b=-3, 2g=-19, 2f=-5$. Thus,$a=4, h=\frac{11}{2}, b=-3, g=-\frac{19}{2}, f=-\frac{5}{2}$. Calculating $2(a-h+b-g+f-12) = 2(4-\frac{11}{2}-3+\frac{19}{2}-\frac{5}{2}-12) = 2(1-\frac{11}{2}+7-\frac{5}{2}) = 2(8-\frac{16}{2}) = 2(8-8) = 0$. Wait,re-evaluating the expansion: $(x+3y-4)(4x-y-3) = 4x^2-xy-3x+12xy-3y^2-9y-16x+4y+12 = 4x^2+11xy-3y^2-19x-5y+12=0$. $2(a-h+b-g+f-12) = 2(4 - 5.5 - 3 + 9.5 - 2.5 - 12) = 2(-1.5 - 3 + 9.5 - 2.5 - 12) = 2(-19/2) = -19$.

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

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