Lines

Answer :

Line through (1, 2) is given by

                             y – 2 = m(x – 1)

                   Let it meet 3x + 4y – 12 = 0 and 3x + 4y – 24 = 0 in point A & B.

                   ∴       Solving lines, y = 2 + mx – m and 3x + 4y – 24 = 0 we get,

                             4(2 + mx – m) = 24 – 3x

                   ∴       8 + 4mx – 4m = 24 – 3x

                   ∴       (4m + 3)x = 16 + 4m

                   ∴       x = [(16 + 4m) / (4m + 3)]

                   ∴       y = [(6 + 21m) / (3 + 4m)]

                   ∴       B = [{(16 + 4m) / (4m + 3)}, {(6 + 21m) / (3 + 4m)}]

                   also   A = [{(4 + 4m) / (3 + 4m)}, {(6 + 9m) / (3 + 4m)}]

                   Now  AB = 3 ⇒ AB² = 9

                   ∴       [(16 – 4) / (3 + 4m)]² = [(12m) / (3 + 4m)]² = 9

                   ∴       [{(12)² + (12)² ∙ m²} / (3 + 4m)²] = 9

                   ∴       m = (7 / 24)

                   ∴       line is y – 2 = (7 / 24)(x – 1)

                   ∴       7x – 24y + 41 = 0

Answer :

y = (x/2) ⇒ slope = m₁ = tan^–1(1/2)

                   y = 3x ⇒ slope = m₂ = tan^–1(3)

                   given:          (a, a²)lies inside angle between lines at the origin

                   ∴ Angle subtended by this point at origin will lie between m₁ & m₂

                   ∴       m = [(a² – 0) / (a – 0)] = a.

                   ∴       tan^–1(1/2) < a < tan^–1(3)

                   i.e.     a ∈ [tan^–1(1/2), tan^–1(3)]

Answer :

P(– 1, 0), Q(0, 0), R(3, 3√3)

                   slope of QR = [(3√3 – 0) / (3 – 0)] = √3

                   i.e.     tanθ = √3 hence θ = (π/3)

                   Let QS be bisector of ∠PQR which passes through (0, 0) and slope = – √3.

                   ∴       eqⁿ of QS =          y – y₁ = m(x – x₁)

                   ∴       y – 0 = – √3 (x – 0)

                   ∴       y = – √3 x

                   ∴       y + √3x = 0

                   i.e.     √3x + y = 0

Answer :

(1/a), (1/b), (1/c) are in H.P.

                   ⇒       (1/a) + (1/c) = (2/b)

                   i.e.     (1/a) – (2/b) + (1/c) = 0

                   line: (x/a) + (y/b) + (1/c) = 0 hence by comparing, x = 1, y = – 2

                   ∴ line passes through (x, y) i.e. (1, – 2)

Answer :

Perpendicular distance between (0, 0) and line 2x + y + 6 = 0 is OQ and

                   OQ = [(|2x + y + 6|) / √(2² + 1²)] when x = 0 and y = 0

                   ∴       OQ = (6 / √5)

                   similarly distance between (0, 0) and line 4x + 2y – 9 = 0 OP and

                   OP = [9 / (2√5)]

                   ratio = (OP / OQ) = (9/2) × (1/6) = (3/4)