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If the intersection of the line y = KX + 2K + 1 and the line y = − 12x + 2 is in the first quadrant, then the value range of K is () A. −12＜k＜12B. −16＜k＜12C. k＞12D. k＞−12
The intersection of two straight lines is: y = KX + 2K + 1y = − 12x + 2, the solution of the equations is: x = 2 − 4k2k + 1y = 6K + 12K + 1, ∵ the intersection of the straight line y = KX + 2K + 1 and the straight line y = − 12x + 2 is in the first quadrant, ∵ 2 − 4k2k + 1 ＞ 06k + 12K + 1 ＞ 0, the solution of the inequalities is: − 16 ＜ K ＜ 12, so choose B
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