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In Column I, a family of concurrent lines is given and their points of concurrence are given in Column II. Match the items of Column I with those of Column I with those of Column II.(A) If a, b, c are real as 2a + 3b + c = 0, then the lines ax + by + c = 0 are concurrent(P) (2/5,3/5)(B) If a is a parameter, then the family of lines (1 + a)x + (2 - a) y + 5 = 0 is concurrent at(q) (2, 3)(C) The lines (a d)x + ay + (a + d) = 0 for different values of d are concurrent at(r) (1, 2)(D) For different values m and n, the lines (m + 2n)x + (m - 3n)y m + n = 0 are concurrent at(s) (-5/3,-5/3)(t) ()1,-2 |
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Answer» (A) 2a + 3b + c = 0 ⇒ ax + by + c = 0 passes through (2, 3). Answer: (A) → (q) (B) The equation (1 + a)x + (2 - a)y + 5 = 0 is written as (x + 2y + 5) + a(x - y) = 0, Hence, the line passes through the intersection of the lines x + 2y + 5 = 0 and x - y = 0 and the point of intersection is (-5/3,-5/3) Answer: (B) → (s) (C) The equation (a - d)x + ay + (a + d) = 0 is written as a(x + y + 1) + d (1 - x) = 0 so that this line passes through the intersection of the lines 1 - x = 0 and x + y + 1 = 0 which is given by (1, -2). Answer: (C) → (t) (D) The equation (m + 2n)x + (m - 3n)y - m + n = 0 is written as m(x + y -1 = 0) . Hence, the line passes through the intersection of the lines x + y - 1 = 0 and 2x - 3y + 1 = 0 which is given by (2/5, 3/5). Answer: (C) → (p) |
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