Question 128Asked in BPSC TRE 2.0 (2023) Mathematics Class 9-10

If the points A(a², 0), B(0, b²) and C(1, 1) are collinear, thenयदि बिंदु A(a², 0), B(0, b²) और C(1, 1) संरेखी हैं, तो

यदि बिंदु A(a², 0), B(0, b²) और C(1, 1) संरेखी हैं, तो
  1. A1/a + 1/b = 11/a + 1/b = 1
  2. Ba² + b² = 1a² + b² = 1
  3. C1/a² + 1/b² = 11/a² + 1/b² = 1
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
Show answerHide answer

Correct answer: (C) 1/a² + 1/b² = 11/a² + 1/b² = 1

Explanation

Collinearity gives zero area: a²(b² - 1) + 0 + 1(0 - b²) = 0, i.e. a²b² = a² + b², so 1/a² + 1/b² = 1.संरेखता से क्षेत्रफल शून्य: a²(b² − 1) + 0 + 1(0 − b²) = 0, अर्थात् a²b² = a² + b², अतः 1/a² + 1/b² = 1।

From BPSC TRE 2.0 (2023) Mathematics Class 9-10 — Previous Year Paper

This is question 128 of BPSC TRE 2.0 (2023) Mathematics Class 9-10 — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.