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

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(A) For the lines to be concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} 2 & 1 & -3 \ 3 & 2 & -2 \ k & -3 & -23 \end{v

Vedclass · Accepted Answer

$1/2, 2/3$

Vedclass · Answer

(A) For the lines to be concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} 2 & 1 & -3 \ 3 & 2 & -2 \ k & -3 & -23 \end{vmatrix} = 0$Expanding along the first row:$2(2(-23) - (-2)(-3)) - 1(3(-23) - (-2)(k)) - 3(3(-3) - 2(k)) = 0$$2(-46 - 6) - 1(-69 + 2k) - 3(-9 - 2k) = 0$$2(-52) + 69 - 2k + 27 + 6k = 0$$-104 + 96 + 4k = 0$$4k - 8 = 0 \implies k = 2$Substituting $k = 2$ into the quadratic equation $6x^2 - 7x + k = 0$,we get:$6x^2 - 7x + 2 = 0$$6x^2 - 4x - 3x + 2 = 0$$2x(3x - 2) - 1(3x - 2) = 0$$(2x - 1)(3x - 2) = 0$The roots are $x = 1/2$ and $x = 2/3$.

If the lines $2x + y - 3 = 0$,$3x + 2y - 2 = 0$,and $kx - 3y - 23 = 0$ are concurrent,then the roots of the equation $6x^2 - 7x + k = 0$ are

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