If the lines L_1 equiv 2x + y + 3 = 0,L_2 equ | vedclass

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(D) The lines $L_1, L_2, L_3$ are concurrent,so the determinant of their coefficients is zero:$\begin{vmatrix} 2 & 1 & 3 \ k & 2 & -3 \ 3 & -2 & 1 \

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$\frac{20}{29}$

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(D) The lines $L_1, L_2, L_3$ are concurrent,so the determinant of their coefficients is zero:$\begin{vmatrix} 2 & 1 & 3 \ k & 2 & -3 \ 3 & -2 & 1 \end{vmatrix} = 0$$2(2 - 6) - 1(k + 9) + 3(-2k - 6) = 0$$-8 - k - 9 - 6k - 18 = 0$$-7k - 35 = 0 \Rightarrow k = -5$Thus,$L_2 \equiv -5x + 2y - 3 = 0$,or $5x - 2y + 3 = 0$.The slope of $L_2$ is $m_1 = \frac{5}{2}$.The slope of the line $2x - 5y + 7 = 0$ is $m_2 = \frac{2}{5}$.The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$.$	an 	heta = \left| \frac{\frac{5}{2} - \frac{2}{5}}{1 + (\frac{5}{2})(\frac{2}{5})} ight| = \left| \frac{\frac{25 - 4}{10}}{1 + 1} ight| = \frac{21}{20}$.Since $	an 	heta = \frac{21}{20}$,we have a right triangle with opposite side $21$ and adjacent side $20$. The hypotenuse is $\sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29$.Therefore,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{20}{29}$.

If the lines $L_1 \equiv 2x + y + 3 = 0$,$L_2 \equiv kx + 2y - 3 = 0$,and $L_3 \equiv 3x - 2y + 1 = 0$ are concurrent,then the cosine of the acute angle between the lines $L_2 = 0$ and $2x - 5y + 7 = 0$ is

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