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Passage: Consider the straight line 3x + y + 4 = 0. Answer the following questions. (i) The point on the line 3x + y + 4 = 0 which is equidistant from the points ( 5, 6) and (3, 2) is (A) ( 1, 1)(B) ( 2, 2) (C) ( 3, 5)(D) ( 1 3 , 3)(ii) Equation of the line passing through the point (1, 1) and perpendicular to the given line is(A) x - 3y + 4 = 0 (B) x - 3y + 5 = 0(C) x - 3y - 4 = 0(D) x - 3y + 2 = 0(iii) If the line y + 5 = k(x - 3) is parallel to the given line then the area of the triangle formed by this line and the coordinate axes (in sq. units) is(A) 8/3(B) 16/3(C) 4(D) 5 |
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Answer» Correct option (i) (B),(ii) (D),(iii) (A) Explanation : (i) Let A = (5 ,6) and B = (3, 2). The slope of AB is and the midpoint of AB = (-1, 4). Hence, the perpendicular bisector of the segment (bar)AB is y - 4 = 2(x + 1) or 2x - y + 6 = 0 . Solving this equation and the given line equations, we have x = -2 and y = 2. Thus, (-2, 2) is the point on the given line which is equidistant from both A(-5 6) and B(3, 2). (ii) Line perpendicular to the given line is of the form y = 1/3x + c This line passes through (1, 1). It implies that 1 = 1/3 + c ⇒ c = 2/3 Thus, the required line is y = x/3 + 2/3 or x - 3y + 2 = 0 (iii) The line y + 5 = k(n - 3) is parallel to the given line ⇒ k = -3. That is, 3x + y = 4 or x/(4/3) + y/4 = 1 Hence, the area of the triangle is 1/2(4/3)(4) = 8/3 |
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