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The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is(a) (1, 1) (b) (2, 2) (c) (–2, –2) (d) (3, 3) |
Answer» (b) (2, 2) Let the point be P whose abscissa = ordinate = a. ∴ P ≡ (a, a) Given, PA = PB ⇒ (a + 1)2 + a2 = a2 + (a – 5)2 ⇒ 2a2 + 2a + 1 = 2a2 – 10a + 25 ⇒ 12a = 24 ⇒ a = 2. ∴ The point is (2, 2). |
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