Question 34Asked in BPSC TRE 2.0 (2023) English Class 6-8 + 2 more papers

Find the points on y-axis which is equidistant from the points (–5, 2) and (9, –2).y-अक्ष पर उन बिन्दुओं को खोजिए, जो बिन्दु (–5, 2) और (9, –2) से समान दूरी पर हों।

y-अक्ष पर उन बिन्दुओं को खोजिए, जो बिन्दु (–5, 2) और (9, –2) से समान दूरी पर हों।
  1. A(0, –5)(0, –5)
  2. B(–7, 0)(–7, 0)
  3. C(0, –7)(0, –7)
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
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Correct answer: (C) (0, –7)(0, –7)

Explanation

Point (0, y): 25 + (y-2)^2 = 81 + (y+2)^2 gives -8y = 56, so y = -7 and the point is (0, -7).बिन्दु (0, y): 25 + (y-2)^2 = 81 + (y+2)^2 से y = -7, अतः बिन्दु (0, -7) है।

From BPSC TRE 2.0 (2023) English Class 6-8 — Previous Year Paper

This is question 34 of BPSC TRE 2.0 (2023) English Class 6-8 — 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.

Asked in 3 papers