The combined equation of two lines L and L_1 | vedclass

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(A) Let the lines be $L: y=mx$,$L_1: y=k_1x$,and $L_2: y=k_2x$. Since $L_1 \perp L_2$,we have $k_1k_2 = -1$,or $k_2 = -\frac{1}{k_1}$.The combined equ

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$26$

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(A) Let the lines be $L: y=mx$,$L_1: y=k_1x$,and $L_2: y=k_2x$. Since $L_1 \perp L_2$,we have $k_1k_2 = -1$,or $k_2 = -\frac{1}{k_1}$.The combined equation of $L$ and $L_1$ is $(y-mx)(y-k_1x) = y^2 - (m+k_1)xy + mk_1x^2 = 0$. Comparing this with $2x^2+axy+3y^2=0$,we rewrite the given equation as $y^2 + \frac{a}{3}xy + \frac{2}{3}x^2 = 0$. Thus,$mk_1 = \frac{2}{3}$ and $-(m+k_1) = \frac{a}{3}$.The combined equation of $L$ and $L_2$ is $(y-mx)(y-k_2x) = y^2 - (m+k_2)xy + mk_2x^2 = 0$. Comparing this with $2x^2+bxy-3y^2=0$,we rewrite the given equation as $y^2 - \frac{b}{3}xy - \frac{2}{3}x^2 = 0$. Thus,$mk_2 = -\frac{2}{3}$ and $-(m+k_2) = -\frac{b}{3}$.We have $k_1 = \frac{2}{3m}$ and $k_2 = -\frac{2}{3m}$. Since $k_1k_2 = -1$,we get $(\frac{2}{3m})(-\frac{2}{3m}) = -1$ $\Rightarrow \frac{4}{9m^2} = 1$ $\Rightarrow m^2 = \frac{4}{9}$ $\Rightarrow m = \pm \frac{2}{3}$.Case $1$: $m = \frac{2}{3}$. Then $k_1 = \frac{2/3}{2/3} = 1$ and $k_2 = -\frac{2/3}{2/3} = -1$. Then $a = -3(m+k_1) = -3(\frac{2}{3}+1) = -3(\frac{5}{3}) = -5$.And $b = 3(m+k_2) = 3(\frac{2}{3}-1) = 3(-\frac{1}{3}) = -1$.Thus,$a^2+b^2 = (-5)^2 + (-1)^2 = 25+1 = 26$.Case $2$: $m = -\frac{2}{3}$. Then $k_1 = \frac{2/3}{-2/3} = -1$ and $k_2 = -\frac{2/3}{-2/3} = 1$. Then $a = -3(m+k_1) = -3(-\frac{2}{3}-1) = -3(-\frac{5}{3}) = 5$.And $b = 3(m+k_2) = 3(-\frac{2}{3}+1) = 3(\frac{1}{3}) = 1$.Thus,$a^2+b^2 = (5)^2 + (1)^2 = 25+1 = 26$.

The combined equation of two lines $L$ and $L_1$ is $2x^2+axy+3y^2=0$ and the combined equation of two lines $L$ and $L_2$ is $2x^2+bxy-3y^2=0$. If $L_1$ and $L_2$ are perpendicular,then $a^2+b^2=$

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