If the lines 4x + 3y - k = 0,2x + y + 3 = 0,a | vedclass

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(D) For the lines to be concurrent,the determinant of their coefficients must be zero: $\begin{vmatrix} 4 & 3 & -k \ 2 & 1 & 3 \ 3 & 2 & k \end{vmat

Vedclass · Accepted Answer

$3$

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(D) For the lines to be concurrent,the determinant of their coefficients must be zero: $\begin{vmatrix} 4 & 3 & -k \ 2 & 1 & 3 \ 3 & 2 & k \end{vmatrix} = 0$ Expanding the determinant: $4(k - 6) - 3(2k - 9) - k(4 - 3) = 0$ $4k - 24 - 6k + 27 - k = 0$ $-3k + 3 = 0 \Rightarrow k = 1$ Substituting $k = 1$ into the equations: $L_1: 4x + 3y - 1 = 0$ $L_2: 2x + y + 3 = 0$ Solving these two equations: $L_1 - 2 	imes L_2$ $\Rightarrow (4x + 3y - 1) - (4x + 2y + 6) = 0$ $\Rightarrow y - 7 = 0$ $\Rightarrow y = 7$ Substituting $y = 7$ into $L_2$: $2x + 7 + 3 = 0$ $\Rightarrow 2x = -10$ $\Rightarrow x = -5$ The point of concurrency is $(-5, 7)$. The perpendicular distance $d$ from $(-5, 7)$ to $3x + 4y + 2 = 0$ is: $d = \left| \frac{3(-5) + 4(7) + 2}{\sqrt{3^2 + 4^2}} ight| = \left| \frac{-15 + 28 + 2}{5} ight| = \left| \frac{15}{5} ight| = 3$.

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