Point P divides the line segment joining R (āˆ’1, 3) and S (9, 8) in ratio k:1. If P lies on the line x – y + 2 = 0, then value of k is
(a) 2/3Ā  (b) 1/2 Ā  (c) 1/3Ā  Ā  (d) 1/4

Ques 35 (MCQ) - Point P divides line segment joining R(-1, 3), S(9,8) - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]

part 2 - Question 35 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 35 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 35 Point P divides the line segment joining R (āˆ’1, 3) and S (9, 8) in ratio k:1. If P lies on the line x – y + 2 = 0, then value of k is (a) 2/3 (b) 1/2 (c) 1/3 (d) 1/4 Using section formula x = (š’ŽšŸ š’™šŸ + š’ŽšŸ š’™šŸ)/(š’ŽšŸ + š’ŽšŸ) x = (š‘˜ Ɨ 9 + 1 Ɨ (āˆ’1))/(š‘˜ + 1) x = (šŸ—š’Œ āˆ’ šŸ)/(š’Œ + šŸ) y = (š’ŽšŸ š’ššŸ + š’ŽšŸ š’ššŸ)/(š’ŽšŸ + š’ŽšŸ) y = (š‘˜ Ɨ 8 + 1 Ɨ 3)/(š‘˜ + 1) y = (šŸ–š’Œ + šŸ‘)/(š’Œ + šŸ) Since point P lies on line It will satisfy it’s equation Putting values of x and y in equation x āˆ’ y + 2 = 0 (šŸ—š’Œ āˆ’ šŸ)/(š’Œ + šŸ)āˆ’(šŸ–š’Œ + šŸ‘)/(š’Œ + šŸ) + 2 = 0 ((9š‘˜ āˆ’1) āˆ’ (8š‘˜ + 3))/(š‘˜ + 1) + 2 = 0 (9š‘˜ āˆ’ 1 āˆ’ 8š‘˜ āˆ’ 3)/(š‘˜ + 1) + 2 = 0 (š‘˜ āˆ’ 4)/(š‘˜ + 1) + 2 = 0 (š‘˜ āˆ’ 4 + 2(š‘˜ + 1))/(š‘˜ + 1) = 0 (šŸ‘š’Œ āˆ’ šŸ)/(š’Œ + šŸ) = 0 3k āˆ’ 2 = 0 3k = 2 k = šŸ/šŸ‘ So, the correct answer is (a)

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