Straight Line

Problem:   A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle $\large x^2 + y^2 = 4 $ . The bisector of angle C belongs to either of two families of concurrent lines, whose points of concurrency are

(A) (√2 , √2 )

(B) (-√2 , -√2 )

(C) (√2 , -√2 )

(D) (-√2 , √2 )

Ans:    (A), (B).

Sol: The points of concurrency lie on the intersection of perpendicular bisectors of AB with the circle x² + y² = 4.

Hence the points are (√2 , √2 )   or (-√2 , -√2 )

Problem: $\large \frac{x}{a} + \frac{y}{b} = 1 $ and $\large \frac{x}{c} + \frac{y}{d} = 1 $ intersect the axes at four concyclic points and c² + a² = b² + d², then these lines can intersect at (a, b, c, d > 0)

(A) (1, 1)

(B) (1, – 1)

(C) (2, – 2)

(D) (3, – 2)

Ans: (A) , (B), (C)

Sol: Points A, B, C, D are concyclic, then ac = bd.

The co-ordinate of the points of intersection of lines are

$\large (\frac{ac(b-d)}{bc-ad}) , (\frac{bd(c-a)}{bc-ad}) $

given c² + a² = b² + d²    ( ac = bd)

⇒ (c – a)² = (b – d)²

⇒ c – a = ± (b – d)

then the locus of the points of intersection is y = ± x.

Problem:   If x + y = 0 is the angle bisector of the angle containing the point (1, 0), for the line  3x + 4y + b = 0, 4x + 3y – b = 0 then b can be

(A) 3

(B) 4

(C) 5

(D) 6

Ans: (C) , (D) 

Sol: (3x + 4y + b) = ± (4x + 3y – b)

For x + y = 0, we have to choose -ve sign.

(3x₁ + 4y₁ + b) (4x₁ + 3y₁ – b) < 0

⇒ (3 + b) (4 – b) < 0

⇒ b > 4 or b < -3.

Problem: ABCD is a square. E(4, 3), F(2, 5) lie on AB and CD respectively such that EF divides the square in two equal parts. If the coordinates of A is (7, 3), other coordinates of the vertices can be

(A) (7, 2)

(B) (7, 5)

(C) (-1, 3)

(D) (-1, 7)

Ans: (B), (C).

Sol: Mid-point (G) of EF is centre of square.

Problem: The minimum distance of $\large 4x^2 + y^2 + 4x – 4y + 5 = 0 $ from the line $-4x + 3 y = 3 $ is

(A) 2

(B) 3

(C) 1

(D) none of these

Ans: (C)

Sol: The given curve represents the point (-1/2 , 2) , so minimum distance = 1.