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JEE MainMathematicsStraight Lines

Consider the three lines L₁: 2x - 3y + 5 = 0 , L₂: 3x + 4y - 1 = 0 , and L₃: kx - 6y + 7 = 0 . If k belongs to the set 1, 2, 3, , 10 , then the sum of all values of k for which the three lines form a triangle is ________.

Correct answer

50

Step-by-step solution

The given lines are: L₁: 2x - 3y + 5 = 0 L₂: 3x + 4y - 1 = 0 L₃: kx - 6y + 7 = 0 First, find the point of intersection of L₁ and L₂ . Multiplying L₁ by 4 and L₂ by 3 and adding them gives: 8x - 12y + 20 = 0 9x + 12y - 3 = 0 17x + 17 = 0 x = -1 Substituting x = -1 in L₁ , we get 2(-1) - 3y + 5 = 0 y = 1 . The intersection point is (-1, 1) . For the lines to NOT form a triangle, L₃ must either pass through (-1, 1) or be parallel to L₁ or L₂ . Case 1: Concurrency L₃ passes through (-1, 1) : k(-1) - 6(1) + 7 = 0 -k + 1

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