11 + Interview Questions in STRAIGHT LINE IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Consider the circles S1 : x2 + y2 + 2x + 4y + 1 = 0 S2 : x2 + y2 – 4x + 3 = 0 S3 : x2 + y2 + 6y + 5 = 0 Which of this following statements are correct? (A) Radical centre of S1, S2 and S3 lies in 1st quadrant. (B) Radical centre of S1, S2 and S3 lies in 4th quadrants. (C) Radius of the circle intersecting S1, S2 and S3 orthogonally is 1. (D) Circle orthogonal to S1, S2 and S3 has its x and y intercept equal to zero.
Answer» radical centre is (1, –1) radius of the circle orthogonally to S₁ , S₂ , S₃is 1 and has the equation x²+ y² – 2x + 2y + 1 = 0

1.

Consider the circles S1 : x2 + y2 + 2x + 4y + 1 = 0 S2 : x2 + y2 – 4x + 3 = 0 S3 : x2 + y2 + 6y + 5 = 0 Which of this following statements are correct? (A) Radical centre of S1, S2 and S3 lies in 1st quadrant. (B) Radical centre of S1, S2 and S3 lies in 4th quadrants. (C) Radius of the circle intersecting S1, S2 and S3 orthogonally is 1. (D) Circle orthogonal to S1, S2 and S3 has its x and y intercept equal to zero.

Answer»

radical centre is (1, –1) radius of the circle orthogonally to S₁ , S₂ , S₃is 1 and has the equation

x²+ y² – 2x + 2y + 1 = 0

Discussion

2.	Given the family of lines, a (3x + 4y + 6) + b (x + y + 2) = 0 . The line of the family situated at the greatest distance from the point P (2, 3) has equation :(A) 4x + 3y + 8 = 0 (B) 5x + 3y + 10 = 0 (C) 15x + 8y + 30 = 0 (D) none
Answer» (A) 4x + 3y + 8 = 0 point of intersection is A (- 2, 0) . The required line will be one which passes through (- 2, 0) and is perpendicular to the line joining (- 2, 0) and (2, 3) or taking (2, 3) as centre and radius equal to PA draw a circle, the required line will be a tangent to the circle at (- 2, 0)

2.

Given the family of lines, a (3x + 4y + 6) + b (x + y + 2) = 0 . The line of the family situated at the greatest distance from the point P (2, 3) has equation :(A) 4x + 3y + 8 = 0 (B) 5x + 3y + 10 = 0 (C) 15x + 8y + 30 = 0 (D) none

Answer»

(A) 4x + 3y + 8 = 0

point of intersection is A (- 2, 0) . The required line will be one which passes through (- 2, 0) and is perpendicular to the line joining (- 2, 0) and (2, 3) or taking (2, 3) as centre and radius equal to PA draw a circle, the required line will be a tangent to the circle at (- 2, 0)

Discussion

3.	A and B are two fixed points whose co-ordinates are (3, 2) and (5, 4) respectively. The co-ordinates of a point P if ABP is an equilateral triangle, is/are :(A) (-4√3 , 3+ √3)(B) (4+ √3 , 3- √3)(C) (3-√3 , 4+√3)(D) (3+ √3 , 4-√3)
Answer» (A) (-4√3 , 3+ √3) (B) (4+ √3 , 3- √3) [ Hint : use parametric ]

3.

A and B are two fixed points whose co-ordinates are (3, 2) and (5, 4) respectively. The co-ordinates of a point P if ABP is an equilateral triangle, is/are :(A) (-4√3 , 3+ √3)(B) (4+ √3 , 3- √3)(C) (3-√3 , 4+√3)(D) (3+ √3 , 4-√3)

Answer»

(A) (-4√3 , 3+ √3)

(B) (4+ √3 , 3- √3)

[ Hint : use parametric ]

Discussion

4.	The straight lines x + y = 0, 3x + y - 4 = 0 and x + 3y - 4 = 0 form a triangle which is (A) isosceles (B) right angled (C) obtuse angled (D) equilateral
Answer» (A) isosceles (C) obtuse angled

4.

The straight lines x + y = 0, 3x + y - 4 = 0 and x + 3y - 4 = 0 form a triangle which is (A) isosceles (B) right angled (C) obtuse angled (D) equilateral

Answer»

(A) isosceles

(C) obtuse angled

Discussion

5.	The locus of the centers of the circles which cut the circles x2 + y2 + 4x - 6y + 9 = 0 and x2 + y2 - 5x + 4y - 2 = 0 orthogonally is : (A) 9x + 10y - 7 = 0 (B) x - y + 2 = 0 (C) 9x - 10y + 11 = 0 (D) 9x + 10y + 7 = 0
Answer» (C) 9x - 10y + 11 = 0 Let out circle be x²+ y² + 2gx + 2fy + c = 0 conditions 2(– g) (–2) + 2( – f ) (3) = c + 9 and 2(– g) (5/2) + 2( – f ) (–2) = c – 2 ∴ ag – 10 f = 11 ∴ locus of centre 9x – 10y + 11 = 0

5.

The locus of the centers of the circles which cut the circles x2 + y2 + 4x - 6y + 9 = 0 and x2 + y2 - 5x + 4y - 2 = 0 orthogonally is : (A) 9x + 10y - 7 = 0 (B) x - y + 2 = 0 (C) 9x - 10y + 11 = 0 (D) 9x + 10y + 7 = 0

Answer»

(C) 9x - 10y + 11 = 0

Let out circle be x²+ y² + 2gx + 2fy + c = 0

conditions 2(– g) (–2) + 2( – f ) (3) = c + 9

and 2(– g) (5/2) + 2( – f ) (–2) = c – 2

∴ ag – 10 f = 11

∴ locus of centre 9x – 10y + 11 = 0

Discussion

6.	Straight lines 2x + y = 5 and x - 2y = 3 intersect at the point A . Points B and C are chosen on these two lines such that AB = AC . Then the equation of a line BC passing through the point (2, 3) is (A) 3x - y - 3 = 0 (B) x + 3y - 11 = 0 (C) 3x + y - 9 = 0 (D) x - 3y + 7 = 0
Answer» (A) 3x - y - 3 = 0 (B) x + 3y - 11 = 0 [Hint: Note that the lines are perpendicular . Find the equation of the lines through (2, 3) and parallel to the bisectors of the given lines, the slopes of the bisectors being - 1/3 and 3 ]

6.

Straight lines 2x + y = 5 and x - 2y = 3 intersect at the point A . Points B and C are chosen on these two lines such that AB = AC . Then the equation of a line BC passing through the point (2, 3) is (A) 3x - y - 3 = 0 (B) x + 3y - 11 = 0 (C) 3x + y - 9 = 0 (D) x - 3y + 7 = 0

Answer»

(A) 3x - y - 3 = 0

(B) x + 3y - 11 = 0

[Hint: Note that the lines are perpendicular . Find the equation of the lines through (2, 3) and parallel to the bisectors of the given lines, the slopes of the bisectors being - 1/3 and 3 ]

Discussion

7.	Column-IColumn-II(A) Two intersecting circles(P) have a common tangent(B) Two circles touching each other(Q) have a common normal(C) Two non-concentric circles, one strictly inside the other(R) do not have a common normal(D) Two concentric circles of different radii(S) do not have a radical axis.
Answer» [ (A) P, Q; (B) P, Q; (C) Q; (D) Q, S]

Discussion

8.	The locus of the middle points of the system of chords of the circle x2 + y2 = 16 which are parallel to the line 2y = 4x + 5 is (A) x = 2y (B) x + 2y = 0 (C) y + 2x = 0 (4) y = 2x
Answer» (B) x + 2y = 0

Discussion

9.	Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.
Answer» 4(2) -7(1) = -7 + 1 = 2 > 0 4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0 Since the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0

9.

Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.

Answer»

4(2) -7(1) = -7 + 1 = 2 > 0

4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0

Since the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0

Discussion

10.	The points (x1, y1) , (x2, y2) , (x1, y2) and (x2, y1) are always : (A) collinear (B) concyclic (C) vertices of a square (D) vertices of a rhombus
Answer» (B) concyclic All the points lie on the circle (x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0 Note that points form a rectangle

10.

The points (x1, y1) , (x2, y2) , (x1, y2) and (x2, y1) are always : (A) collinear (B) concyclic (C) vertices of a square (D) vertices of a rhombus

Answer»

(B) concyclic

All the points lie on the circle (x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0

Note that points form a rectangle

Discussion

11.	B and C are points in the xy plane such that A(1, 2) : B (5, 6) and AC = 3BC. Then (A) ABC is a unique triangle (B) There can be only two such triangles. (C) No such triangle is possible (D) There can be infinite number of such triangles.
Answer» (D) There can be infinite number of such triangles. Locus of C is a circle 8(x²+ y²) – 88x – 104y + 544 = 0 ⇒ (D)

11.

B and C are points in the xy plane such that A(1, 2) : B (5, 6) and AC = 3BC. Then (A) ABC is a unique triangle (B) There can be only two such triangles. (C) No such triangle is possible (D) There can be infinite number of such triangles.

Answer»

(D) There can be infinite number of such triangles.

Locus of C is a circle 8(x²+ y²) – 88x – 104y + 544 = 0 ⇒ (D)

Discussion

Explore topic-wise InterviewSolutions in .

Given the family of lines, a (3x + 4y + 6) + b (x + y + 2) = 0 . The line of the family situated at the greatest distance from the point P (2, 3) has equation :(A) 4x + 3y + 8 = 0 (B) 5x + 3y + 10 = 0 (C) 15x + 8y + 30 = 0 (D) none

A and B are two fixed points whose co-ordinates are (3, 2) and (5, 4) respectively. The co-ordinates of a point P if ABP is an equilateral triangle, is/are :(A) (-4√3 , 3+ √3)(B) (4+ √3 , 3- √3)(C) (3-√3 , 4+√3)(D) (3+ √3 , 4-√3)

The straight lines x + y = 0, 3x + y - 4 = 0 and x + 3y - 4 = 0 form a triangle which is (A) isosceles (B) right angled (C) obtuse angled (D) equilateral

The locus of the centers of the circles which cut the circles x2 + y2 + 4x - 6y + 9 = 0 and x2 + y2 - 5x + 4y - 2 = 0 orthogonally is : (A) 9x + 10y - 7 = 0 (B) x - y + 2 = 0 (C) 9x - 10y + 11 = 0 (D) 9x + 10y + 7 = 0

Straight lines 2x + y = 5 and x - 2y = 3 intersect at the point A . Points B and C are chosen on these two lines such that AB = AC . Then the equation of a line BC passing through the point (2, 3) is (A) 3x - y - 3 = 0 (B) x + 3y - 11 = 0 (C) 3x + y - 9 = 0 (D) x - 3y + 7 = 0

Column-IColumn-II(A) Two intersecting circles(P) have a common tangent(B) Two circles touching each other(Q) have a common normal(C) Two non-concentric circles, one strictly inside the other(R) do not have a common normal(D) Two concentric circles of different radii(S) do not have a radical axis.

The locus of the middle points of the system of chords of the circle x2 + y2 = 16 which are parallel to the line 2y = 4x + 5 is (A) x = 2y (B) x + 2y = 0 (C) y + 2x = 0 (4) y = 2x

Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.

The points (x1, y1) , (x2, y2) , (x1, y2) and (x2, y1) are always : (A) collinear (B) concyclic (C) vertices of a square (D) vertices of a rhombus

B and C are points in the xy plane such that A(1, 2) : B (5, 6) and AC = 3BC. Then (A) ABC is a unique triangle (B) There can be only two such triangles. (C) No such triangle is possible (D) There can be infinite number of such triangles.