If the lines x+y-1=0,kx+2y+1=0,and 4x+2ky+7=0 | vedclass

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Question

(C) For the lines to be concurrent,the determinant of their coefficients must be zero:$\left|\begin{array}{ccc}1 & 1 & -1 \ k & 2 & 1 \ 4 & 2k & 7\e

Vedclass · Accepted Answer

$\frac{-13}{2}$

Vedclass · Answer

(C) For the lines to be concurrent,the determinant of their coefficients must be zero:$\left|\begin{array}{ccc}1 & 1 & -1 \ k & 2 & 1 \ 4 & 2k & 7\end{array}ight|=0$Expanding the determinant:$1(14 - 2k) - 1(7k - 4) - 1(2k^2 - 8) = 0$$14 - 2k - 7k + 4 - 2k^2 + 8 = 0$$-2k^2 - 9k + 26 = 0$$2k^2 + 9k - 26 = 0$Factoring the quadratic equation:$2k^2 + 13k - 4k - 26 = 0$$k(2k + 13) - 2(2k + 13) = 0$$(2k + 13)(k - 2) = 0$So,$k = 2$ or $k = -\frac{13}{2}$.If $k = 2$,the lines are $x+y-1=0$,$2x+2y+1=0$,and $4x+4y+7=0$. The first two lines are parallel ($x+y-1=0$ and $x+y+0.5=0$),so they cannot be concurrent.Therefore,the only valid value is $k = -\frac{13}{2}$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

If the lines $x+y-1=0$,$kx+2y+1=0$,and $4x+2ky+7=0$ are concurrent,then $k=$

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