Question

Passage: Consider the family of concurrent lines which are concurrent at (1, 2) represented by the equation (3x – y – 1) λ (4x – y – 2) = 0, where L is a parameter. Answer the following questions.

(i) A member of the family with positive slope which makes an angle 45° with the line (3x - 4y - 2) = 0 is

(A)  7x - y - 5 = 0

(B)  4x - 3y + 2 = 0

(C)  x + 7y - 15 = 0

(D)  5x - 3y - 4 = 0

(ii)  Equation of the line belonging to the given family which is perpendicular to the line x + y - 1 = 0  is

(A)  x - y + 1 = 0

(B)  x +  y - 3 = 0

(C)  2x + y -4 = 0

(D)  3x - y - 1 = 10

(iii) The locus of the feet of the perpendiculars from the origin on each of the lines of the members of the family is

(A)  (2x - 1)² + 4(y + 1)² = 5

(B)  (2x - 1)² + (y + 1)² = 5

(C)  (2x + 1)² + 4(y - 1)² = 5

(D)  (2x - 1)² + 4(y - 1)² = 5

Answer 1

Correct option (i) (A),(ii) (A),(iii) (D)

Explanation :

(i) The slope of a line belonging to the given family is

3 + 4λ/1 + λ

and the slope of the line 3x - 4y - 2 = 0 is 3/4. Therefore, by hypothesis, we have

Therefore, 16λ + 13 = ±(13λ + 9). Hence

λ = - 4/3, -22/29

Case 1: When we have

λ = - 4/3

the slope is

Hence, the required equation is

Case 2: When we have

λ = -22/29

the slope is 

3 - (88/29)/1 - (22/29) = -1/7

Thus, the required line is 7x - y - 5 = 0

 (ii) By hypothesis, we have

Hence, the required line is

(iii)  Let A = (1, 2) at which the given family of lines are concurrent and O be the origin. If P is the foot of the perpendicular from origin O onto any line of the family then P lies on the circle drawn on OA as diameter because ΔAPO is equal to 90°. The circle with (0, 0) and (1, 2) as ends of a diameter is