SL

1. A line has intercepts a and b on the coordinate axes. If keeping the origin fixed, the coordinate axes are rotated through 90°, the same line has intercepts p and q, then(A) p =a, q = b (B) p = b, q = a(C) p = -b, q = -a (D) p = b, q = -a

1. (D) Obviously, new x-axis is old y-axis and new y-axis is old (-x) – axis. Thus p =b, q = -a.

2. The equations of the lines representing the sides of a triangle are 3x – 4y = 0, x + y = 0 and 2x – 3y = 7. The line 3x + 2y = 0 always passes through the(A) incentre (B) centroid (C) circumcentre (D) orthocentre

2. (D) 3x + 2y =0 passes through the point of intersection of two lines ( i.e. origin) and is perpendicular to the third.

3. If the mid-points P, Q and R of the sides of the D ABC are (3, 3) , ( 3, 4) and (2, 4) respectively, then D ABC is (A) right angled (B) acute angled (C) obtuse angled (D) isosceles

3. (A), (D) D ABC is similar to D PQR, which is isosceles right triangle.

4. If the line y = x cuts the curve x4 + ax2y + bxy + cx + dy + 6 = 0 at A, B, C and D, then OA.OB.OC.OD ( where O is the origin) is (A) a – 2b +c (B) 2c2d (C) 96 (D) 6

4. (C)

The line y = can be written as x = , y = . If this line cuts the given curve, then

.

Therefore OA. OB.OC.OD = |r1| |r2| |r3| |r4| = |r1 r2 r3 r4| = 96

5. A curve with equation of the form y = ax4 + bx3+cx +d has zero gradient at the point (0, 1) and also touches the x-axis at the point ( -1, 0). Then the values of x for which the curve has negative gradients are(A) x > -1 (B) x < 1(C) x < -1 (D) -1 £ x £ 1

5. (C) y = ax4 + bx3 +cx +d . . . . (1) y touches x-axis at (-1, 0) so, ( -1, 0) lies on it and dy/ dx =0so, 0 = a – b – c +d . . . . (2)From (1) dy/ dx = 4ax3 + 3bx2 +c . . . . (3)

Hence Þ - 4a + 3b +c =0 . . . . (4)

Also = c =0 (since curve touches (0, 1) )

Þ ( 0, 1) also lies on it hence d = 1

Putting values of c and d in (2) and (4), solving for a and b we get a = 3, b = 4

1

Therefore (3) becomes = 12x3 +12x2

Now dy/dx < 0 Þ 12 x3 + 12x2 < 0Þ 12x2 (x+1) < 0 Þ x < -1

6. If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c - 1 = 0 always passes through a fixed point is (A) 0 (B) 20 (C) 30 (D) none of these

6. B

Equation of line has two independent parameters. It can pass through a fixed point if it

contains only one independent parameter. Now there must be one relation between and

independent of a, b and c so that can be expressed in terms of and straight line contains only one

independent parameter. Now that given relation can be expressed as .

Now RHS be independent of c if t = 20.

7. Let y = mx + 5 be a line. Then the range of m so that point (5, 6) and (-6, 6) always lie on the opposite side of origin is

(A) m > and m < - (B) < m <

(C) m > (D) none of these

7. BFor (0, 0)y – mx – 5 = –5 < 0so, for (5, 6) and (–6, 6) to lie on opposite side we have 6 – 5m – 5 > 0 and 6 + 6m – 5 > 0

Þ – < m < .

8. As system of line is given as y = m ix + ci, where mi can take any value out of 0, 1, -1 and when mi is positive then ci can be 1 or -1 when mi equal 0, ci can be 0 or 1 and when mi equals -1, ci can take 0 or 2. Then the area enclosed by all these straight line is

(A) (B)

(C) (D) none of these

8. CLines are y = 1, y = 0 y = -x, y = -x + 2 y = x + 1, y = x - 1 Area of OABCDE = area of OBGF

=

2

9. Two sides of a rhombus OABC ( lying entirely in first quadrant or fourth quadrant ) of area equal to 2 sq. units,

are y = , y = x . Then possible coordinates of B is / are (‘O’ being the origin)

(A) (B) (C) (D) none of these

9. (A), (B)Let the length of the sides of the rhombus be lÞ Area = l2 sin30° =2 Þ l = 2 unitOB2 = OA2 + AB2 – 2OA .AB cos150°

= 4 + 4 – 2 (4). = 4 (2 + )

OB = 2 B º º Hence coordinates of B can be or

10. If DOAB is an equilateral triangle (O is the origin and A is a point on the x-axis), then centroid of the triangle will be (A) always rational (B) rational if B is rational(C) rational if A is rational (D) never rational (a point P(x, y) is said to be rational if both x and y are rational)

10. (D)

Centroid is . Now clearly and cannot be rational simultaneously for any real a.

11. If a – b, b – c, c – a are in A.P., then the straight line (a –b)x + ( b – c)y +(c –a) = 0 will pass through (A) (1, 2) (B) (2, 1)(C) (2, 3) (D) (3, 1)

11. (A)a –b +c – a = 2(b –c)Þ c –b = 2( b –c) Þ c= bÞ a –b = -( c- a) So, equation of straight line is x –1 = 0.

12. If the quadratic equation ax2 + bx + c = 0 has –2 as one of its roots then ax + by + c = 0 represents(A) A family of concurrent lines (B) A family of parallel lines(C) A single line (D) A line perpendicular to x-axis

12. (A)Since – 2 is a root of ax2 + bx + c = 0 Þ 4a – 2b + c = 0So, ax + by + c = 0 always passes through (4, - 2)

13. Two particles start from the same point (2, -1), one moving 2 units along the line x + y = 1 and the other 5 units along x – 2y = 4. If the particles move towards increasing y, then their new position will be(A) (2 - Ö2, Ö2 –1) (B) (2Ö5 + 2, Ö5 – 1)(C) (2 + Ö2, Ö2 + 1) (D) (2Ö5-2, Ö5 –1)

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13. (A), (B)Let P (2, -1) goes 2 units along x + y = 1 upto A and 5 units along x – 2y = 4 up to B

Slope of PA = -1= tan135° , slope of PB = = tanq

\sinq = , cosq =

\ A º (x1 + rcos135°, y1 = rsin135°) = (2 + 2x , -1 + 2 ´ )

= (2 - Ö2, Ö2 – 1)

B º (x1 + rcosq, y1 + rsinq) = (2 + 5 ´ , -1 + ) = (2Ö5 + 2, Ö5 – 1)

14. In an isosceles right angled triangle, a straight line drawn from the mid-point of one of equal sides to the opposite angle . It divides the angle into two parts, q and (p/4 - q). Then tanq and tan[(p/4) -q] are equal to

(A) (B)

(C) (D) none of these

14. (A)Let ABC be the isosceles triangle right angled at A. E is the mid-point of AC and ÐABE = q. B =C = 45°,

AB = AC,AE = EC = AC and

ÐCBE = -q . From the right angled triangle

ABE,

we get tanq =

tan( -q) = .

15. Consider a line passing through ( , 1) and (1, ) then number of rational points lying in the line is (A) 1 (B) at most 2(C) ¥ (D) none of these

15. (D) slope m =

Let there be a rational point P (x, y)

Now slope of line joining ( , 1) and (x, y) is

It is obvious that any rational choice for x, y it is not equal to .

16. A ray of light travelling along the line x + y = 1 is inclined on the x-axis and after refraction it enters the other side of the x-axis by turning 15° away from the x-axis. The equation of the line along which the refraction ray travels is(A) y - x +1 = 0 (B) y + x +1 = 0(C) y + x -1 = 0 (D) none of these .

16. (D)

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The line of the refracted ray passes through the point (1, 0) and its slope is tan120° \ The equation of the line of the refracted ray is y - 0 = tan120° ( x –1)y = - ( x –1) Þ y + x – = 0.

17. If P º ; Q º ; R º where xk ¹ 0, k = p,q, r Î N, denotes the

kth term of a Harmonic progression, then

(A) Area(DPQR) =

(B) DPQR is a right angled triangle (C) the points P, Q and R are collinear. (D) none of these .

17. (C)

consider

= = 0

Þ Points P, Q and R are collinear.

18. If 3l + 5m + 7n = 0 then the family of straight line lx + my + n = 0 passes through the fixed point

(A) (2, 5) (B)

(C) (D)

18. (D) 3l + 5m + 7n = 0

Þ ……(1)

also lx + my + n = 0 ……(2)

\ l

This shows that the family of straight lines passes through the fixed point .

19. The number of points on the line 3 x + 4y = 5, which are at a distance of sec2q +2 cosec2q, q Î R, from the point (1, 3), is (A) 1 (B) 2(C) 3 (D) infinite

19. (B)The perpendicular distance of (1, 3) from the line 3x + 4y = 5 is 2 units while sec2q +2cosec2q ³ 3 (as sec2q , cosec2q ³ 1)Evidently, there will be two such points on the line.

20. If the median AD of a DABC is bisected at E and BE intersects AC in F, then is equal to

(A) (B)

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(C) (D)

20. (B)

Let A be the origin and position vectors of B and C be respectively.

Now , , and

Equation of …(1)

Equation of …(2)

For the point F, , for some particular l and m .

Since are non-collinear,

Þ l = 4/3, m = 1/3

Þ Position vector of F = m = .

Thus .

21. The base BC of DABC is bisected at (p, q) and equation of sides AB and AC are px + qy = 1 and qx + py = 1. Then equation of median through A is(A) (2q – 1) (px + qy) = pq(B) (2pq – 1) (px + qy – 1) = (p2 + q2 – 1) (qx + py – 1)(C) (px + qy – 1) (qx + py – 1) = 0(D) None of these.

21. (B)Any line through A is given by (px + qy – 1) + l(qx + py – 1) = 0 which is passing through (p, q).

Hence l = .

Thus the required line is ( px + qy –1) – .

22. P is a point on either of the two lines y - at a distance of 3 units from their point of intersection. The coordinates of the foot of the perpendicular from P on the bisector of the angle between them are

(A) Either

304

21

,0 or

- 304

21

,0 depending on the location of P.

(B)

(C)

(D) None of these

22. (B)

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The lines intersect at (0, 2). The equation of one of the lines can be written as

= = r. The coordinates of the point at a distance 3 units from (0, 2) . The coordinates

of the point at a distance of 3 units from (0, 2) on the other line are . The required point is the

mid-point of these points i.e. .

23. The straight line x – y + 2 = 0 rotates about a point where it cuts the y-axis and becomes perpendicular to the straight line ax + by + c = 0. In the new position, its equation is(A) ay – bx + 2b = 0 (B) ay – bx – 2a = 0(C) ay – bx + 2a = 0 (D) None of these

23. (B)The given line cuts the y-axis at (0, 2). The equation of the line perpendicular to the given line is bx – ay + k = 0. Since it passes through (0, 2), k = 2a. Hence the required line is bx – ay + 2a = 0.

24. The incentre of the triangle formed by the lines y = |x| and y = 1 is (A) (0, 2 - ) (B) (2 - , 0)(C) (2 + , 0) (D) (0, 2 + )

24. (A) Incentre º (0, 2 - )

Here a = 2 , b = , c = .

25. The area of the square formed by the lines |y| =1 - x and |y| = x +1 equals to

(A) sq. unit (B) sq. unit

(C) 2 sq. unit (D) 1 sq. units

25. (C) Area of the fig. is = 2

26. If the sum of the reciprocals of the intercepts made by a line on the coordinate axes is 1/5, then the line always passes through (A) ( 5, -5) (B) ( -5, 5)(C) ( -5, -5) (D) ( 5, 5)

26. (D) We have = .

Þ Þ line passes through ( 5, 5).

27. Equation of a line passing through the intersection of the lines 2x +y =3 and x + y = 1 and perpendicular to the line y = 2x +k is (A) x - 2y =0 (B) x+ 2y =0(C) y – x =0 (D) y +x = 0

27. (B) Equation of a line passing through intersection of 2x +y – 3 =0 and x +y –1 = 0 is 2x +y –3 + l ( x + y – 1 ) = 0

Þ ( 2+l) x + (1 + l)y – 3 – l = 0 Þ slope = – .

Perpendicularity condition Þ ´ 2 = 1

Þ l = – 3 Therefore, the required equation is - x – 2y = 0 Þ x + 2y = 0

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28. If a – b , b – c, c –a are in A. P. where a, b, c, Î R, then the line ax + by +c = 0 represents (A) family of lines passing through ( 0, -1)(B) family of lines passing through ( -1, 0)(C) general eqution of a line (D) none of these .

28. (A) a – b, b –c, c –a are in A.P. Þ 2 ( b – c)= a –b +c –a Þ b = cTherefore ax +by +c =0 Þ ax +b( y+1) = 0which represents family of lines passing through (0, -1) .

29. If 16a2 +25b2 – c2 = 40ab, then the family of lines ax + by +c = 0 is concurrent at the point(s) (A) (4, -5) (B) (-4, -5)(C) (- 4, 5) (D) none of these

29. (A) , (C)(4a – 5b)2 – c2 = 0 Þ (4a – 5b +c) (4a –5b –c) = 0either 4a –5b + c = 0, or, 4a – 5b – c = 0 Þ – 4a + 5b + c = 0

30. If the intercept made on the line y = mx by the lines x = 2 and x =5 is less then 5, then the range of m is

(A)

-

34,

34

(B)

(C) (D) none of these.

30. (A)The distance between (2, 2m) and ( 5, 5m) is less then 5.Þ ( 5 –2)2 +( 5m – 2m)2 < 25Þ 9m2 < 16

m2 < Þ .

31. The coordinates of the point(s) on the line x + y = 5, which is/are equidistant from the lines |x| = |y|, is/are(A) (5, 0) (B) (0, 5)(C) (-5, 0) (D) (0, -5)

31. (A), (B) |x| = |y| Þ x + y = 0 and x – y = 0. Bisectors of the angles between these are

Þ y = 0 and x = 0. By solving y = 0, x + y = 5,

we get (5, 0) and x =0, x + y = 5, we get ( 0, 5)

32. The equations of three sides of a triangle are x = 5, y – 2 = 0 and x + y = 9. The coordinates of the circumcentre of the triangle are (A) ( 6, 3) (B) (6, -3)(C) ( -6, 3) (D) none of these.

32. (A)Since the given triangle is right angled at (2, 5), therefore the middle point of the hypotenuse is the circumcentre.

33. If in a D ABC (whose circumcentre is origin), a £ sinA, then for any point (x, y) inside the circumcircle of DABC, (A) |xy| < 1/8 (B) |xy| > 1/8(C) 1/8 < xy < 1/2 (D) none of these

33. (A)

a £ sinA Þ Þ R £ 1/2

8

so for any point (x, y) inside the circumcircle, x2 +y2 < 1/4 Þ |xy| < 1/8.

34. P is a point on either of the two lines y – at a distance of 5 units from their point of intersection. The co-ordinates of the foot of perpendicular from P on the bisector ( which passing through the origin) of the angle between them are

(A) (B)

(C) (D) none of these

34. A

AP = 5 units In D AMP, Ð PAM = 30 °

So AM =

Hence coordinates of M is

35. If all the chords of the curve 3x2 – y2 – 2x + 4y = 0, which subtend a right angle at the origin pass through a fixed point, then the coordinates of the fixed point is (A) (1, 1) (B) (2, -1) (C) (-2, 1) (D) (1, -2)

35. D Let lx + my =1 be any chord of the curve . The lines joining the origin to the point of intersection of the chord and the curve are 3x2 – y2 - (2x – 4y) ( lx + my) = 0 lines will be ^ if ( 3 – 2l) + ( – 1 + 4m) = 0 Þ l – 2m = 1 passes through the fixed point (1, -2).

36. The number of lines that can be drawn from the point (2, 3), so that its distance from (-1, 6) is equal to 6, is (A) 1 (B) 2(C) 0 (D) infinite

36. (C)The distance between (2, 3) and ( -1, 6) = 3 . Obviously there can be no line passing through (2, 3) so that its distance from (-1, 6) is 6.

37. The algebraic sum of distances of a variable line from points (2,0), (0,2) and (1,1) is zero. This line passes through a point whose co-ordinates are(A) (1, 1) (B) (1,2)(C) (2,1) (D) None of these

37. (A)Let ax + by +c = 0 be the variable line

9

\

Þ a + b + c = 0Þ (1,1) satisfies ax + by + c = 0

38. If (sin q, cos q); (q Î [0, 2p]) and (1, 4) lie on the same side or on the line x –y + 1 = 0, then maximum value of sin q will be (A) 1 (B) 0

(C) (D) 1 –

38. Bsince .1 –4 + 1 < 0, so

sin q –cos q + 1 £ 0

Þ sin q – cos q £ – Þ sin £ –

Þ Þ £ q £ 2p Þ maximum value of sin q is 0.

39. 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

39. (B)All the four points satisfy the equation of the circle(x-x1) (x-x2) + (y-y1) (y-y2) = 0So the given points are concyclic.

40. Two lines are given by (x-2y)2 +k(x-2y) = 0. The value of k so that distance between them is 3 is(A) k = 0 (B) k = 3Ö5(C) k = -3 (D) k = 3

40. (B)

41. If the lines x = a + m, y = -2 and y = mx are concurrent, the least value of |a| is(A) 0 (B) Ö2(C) 2Ö2 (D) None of these

41. (C)From the given equations, we getm2

+ am + 2 = 0since m is real, a2 ³ 8, |a|³2Ö2So least value of |a| is 2Ö2

42. If all the 3 vertices of an isosceles right angle triangle be integral points and length of base is also an integer, then which of the point is never a rational point. (A point P (x, y) is integer point if both x and y are integers and point is rational if both x and y are rational)(A) centroid (B) in centre(C) circumcentre (D) orthocentre

42. B

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42. Since a is an integer, a is irrational now incentre will be

which is clearly an irrational point.

43. The area of the quadrilateral formed by y = 1 – x, y = 2 – x and the coordinate axes is (A) 1 (B) 2(C) 3 (D) None of these

43. (D)The required area = area of D OBC – area of DOAD

=

44. The set of lines x tan–1a + y sin–1 + 2 = 0 where aÎ(0, 1), are concurrent at;

(A) (B)

(C) (p, p) (D) none of these

44. B

Since a Î (0, 1) Þ sin–1 = cot– 1a =

x tan–1a + y sin–1 + 2 = 0 Þ x tan–1a + y + 2 = 0

Þ (x – y)tan– 1a + = 0

Clearly this passes through .

45. x3 + y3 = 0 represents (A) a point (B) three straight lines (C) combined equation of a straight line and a circle (D) none of these

45. (D)x3 + y3 = 0 Þ x + y = 0 (as x2 + y2 - xy ¹ 0) Þ y = - x

46. If point (a, a) lies in between the lines whose equations are given by x + y = 2, x + y = -2, then (A) |a| = 1 (B) |a| < 1(C) |a| = 2 (D) 2|a| < 1

46. (B)(a, a) lies on y = x, which intersects the given lines at (1, 1) and (-1, -1) respectively. Thus -1 < a < 1. Hence |a| < 1.

47. The orthocentre of the triangle formed by the lines 2x2 + 3xy – 2y2 – 9x + 7y – 5 = 04x + 5y – 3 = 0 lies at(A) ( 3/5 , 11/5) (B) (6/5, 11/5)(B) (5/6, 11/5) None of these

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47. AThe pair of straight lines 2x2 + 3xy – 2y2 – 9 x + 7y– 5 = 0 are perpendicular to each other so orthocentre is point of intersection of these lines

48. Let ax+by+c = 0 be a variable straight line, where a, b and c are 1st, 3rd and 7th term of some increasing A.P. Then the variable straight line always passes through a fixed point which lies on(A) x2+y2 = 13 (B) x2+y2 = 5

(C) (D) 3x+4y=9

48. A, CLet A.P. be l,l + m, l+2m, l+3m,…Given that l = a, l+2m = b, l+6m = cClearly2a-3b+c=0So fixed point is (2,-3)

49. Let ABC be a triangle with equations of the sides AB, BC, and CA respectively be x – 2 = 0y – 5 = 0 and 5x+ 2y – 10= 0 Then orthocentre of the triangle lies on the line (A) x – y = 0 (B) 3x– y = 1(C) 4x + y = 13 (D) x – 2 y = 1

49. B, CThe given triangle is a right triangle, Hence orthcentre is the vertex containing right angle, So orthcentre is (2 , 5) which lies on the lines 3x – y -= 1, and 4x + y = 13

50. Let 2x- 3y = 0 be a given line and P (sinq, 0) and Q(0, cosq) be two points. Then P and Q lies on the same side of the given line if q lies in the (A) I st quadrant (B) II nd quadrant(C) IIIrd quadrant (D) IV th quadrant

50. B, DP and Q lies on the same side if 2 sinq and –3 cosq have same sign i.e. sinq and cosq have opposite sings , which is true in IInd and IVth quadrant

51. The quadratic equation whose roots are the x and y intercepts of the line passing through (1,1) and making a triangle of area A with axes is (A) x2 + Ax + 2A = 0 (B) x2 – 2Ax +2A = 0(C) x2 – Ax + 2A = 0 (D) None of these

51. B

Equation of the line having intercepts a and b on x and y – axis respectively is . Since it passes

through (1, 1), (1)

Since area of triangle formed by the line and axes is A,ab= 2A (2)from (1) and (2) we get a + b = 2A. Thus required quadratic equation is x2 – ( a+b) x+ ab = 0 or x2 – 2Ax + 2A = 0.

52. The difference of the slopes of the lines represented by the equation x2(sec2q - sin2q) - 2xy tanq + y2sin2q = 0 is(A) 2 (B) 2 tan q(C) tan 2q (D) 1

52. x2(sec2q - sin2q) - 2xy tanq + y2sin2q = 0 is;Let it represents y = m1x and y = m2x lines then (y –m1x) (y –m2x) = 0y2 –(m1 + m2) xy + m1m2 = 0

12

Þ m1 + m2 = = 4 cosec 2q

and m1m2 =

= 4 cosec2 2q -1 then (m1 –m2)2 = (4 cosec 2q)2 –4(4 4 cosec2 2q -1)= 164 cosec2 2q -164 cosec2 2q + 4 = 4m1 –m2 = ±2

53. Point P is on the line segment joining (2, 0) and (0, 3) and point Q is an line segment joining (6, 0) and (0, 6). Then the length of the shortest possible segment PQ is (A) 2 (B) 2(C) 2 + (D)

53. BShortest segment PQ = length of perpendicular from (0, 2) on the line joining (6, 0) and (0, 6)= 2

54. Let S be the region consisting of all those points P lying inside the rectangle OABC, where O º (0, 0), A º (3, 0), B º (3, 4) and C (0, 4) which satisfy d (P, OA) £ min {d(P, AB), d(P, BC), d (P, OC)}, where d denotes the distance of P from the corresponding line. Then the area of the region S is (A) 1 (B) 1.5 sq. unit (C) 2.5 (D) none of these

54. B

55. If a, b, c are in A.P. then the straight line ax + by + c = 0 will always pass through a fixed point whose co-ordinates are(A) ( 1, –2) (B) (–1, 2)(C) (1, 2) (D) (–1, –2)

55. ASince a, b,c, are in A.P.Hence 2b = a+c or a – 2b + c =0Hence the given line pass through (1, –2)

56. Locus of the point of intersection of linesx cosa+ y sin a = a and x sin a – y cos a =a (a Î R ) is (A) x2 + y2 =a2 (B) x2 + y2 = 2a2

(C) x2 + y2 + 2x + 2y = a2 (D) none of these

56. BWe have x cosa + y sin a = aAnd x sina - y cos a = aBy squaring and adding (i) and (ii)x2 (cos2a + sin2a) + y2 (sin2a+cos2 a) = a2 +a2

or x2 + y2 = 2a2

57. In a DABC, the mid-points of the sides AB, BC and CA are ( 0, 2) ( 2, 4) and (4, 4) respectively. There exists a point D in the interior of triangle ABC such that Ar(DDAC) = Ar(DDBC) = Ar(DDAB). Then the coordinates of the point D are

(A) (B)

(C) (D) none of these.

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57. (C)

58. If the equations of the three sides of a triangle are 2x + 3y =1, 3x–2y +6 = 0 and x + y =1, then the orthocentre of the triangle lies on the line (A) 13x +13 y = 1 (B) 169x +26 y = -178(C) 169x + y = 0 (D) none of these.

58. (D)

59. The sides of a triangle are x + y = 1, y = 0 and y = x. Then the following is/are the interior point(s) of the triangle. (A) circum centre (B) centriod(C) incentre (D) orthocentre

59. (C)

60. The locus of a point which divides a line segment AB = 4cm in 1 : 2, where A lies on the line y = x and B lies on the y = 2x is(A) 234x2 + 153y2- 378xy – 32 = 0(B) 234x2 + 153y2- 378xy + 32 = 0(C) 234x2 + 153y2 + 378xy + 32 = 0(D) None of these

60. (A)

61. Let I, I1, I2, I3 be the incentre and excentres of the triangle ABC. A triangle is constructed randomly by selecting 3 points from the given points. Then the fourth point would always be the (A) incentre of the formed triangle (B) circumcentre of the formed triangle.(C) orthocentre of the formed triangle.(D) nothing can be said in genernal.

61. (C)

62. The equation of the straight line passing through (4, 0) and parallel to the line 2x - 2y - 3 = 0 is(A) 2x - 2y = 10 (B) 2x - 2y - 8 = 0(C) 2x - 2y + 2 = 0 (D) x - y = 6

62. BLet the line be 2x - 2y = k. This passes through (4, 0)Þ k = 8Þ 2x - 2y - 8 = 0

63. The three lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0, a ¹ b = c, are concurrent if(A) a + b + c = 0 (B) a2 + b2 + c2 = ab + bc + ca(C) a3 + b3 + c3 = abc (D) None of these

63. A

The lines are concurrent if = 0

Þ a3 + b3 + c3 – 3abc = 0

Þ (a + b + c) = 0

Þ a + b + c = 0 (Q a ¹ b = c)

64. The co-ordinates of the foot of the perpendicular from (a, 0) on the line y = mx + a/m are

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(A) (B)

(C) (D) None of these

64. CEquation of the line through (a, 0) and perpendicular to y = mx + a/m is y = -1/m(x - a). Hence the foot of perpendicular is (0, a/m).

65. The angle between the pair of lines given by the equation x2 + 2xy - y2 = 0 is

(A) (B)

(C) (D)

65. AHere, a + b = 1 – 1 = 0 Þ q = p/2

66. If al + bm + cn = 0, then the family of lines lx + my + n = 0 are concurrent at.

(A) (B)

(C) (D) None of these

66. A

We have = 0 and x. + y + 1 = 0. Eliminate l/n.

We get cx – a + (bx – ay) = 0.

This is a family of lines passing through the point of intersection of cx – a = 0 and

bx – ay = 0 Þ x = , y =

67. Area of the quadrilateral formed by lines |x| + |y| = 1 is (A) 4 (B) (C) 2 (D) 1

67. CThe lines are x + y = 1, x – y = 1, – x + y = 1, – x – y = 1. These lines form a square of side Ö2. Area = Ö2.Ö2 = 2.

68. The orthocentre of the triangle formed by the lines xy = 0 and 2x - 3y = 5 is

(A) (B) (1, -1)

(C) origin (D)

68. C

69. In triangle ABC, A(-1, 3), B(1, -1) and C(5, 1) the length of median through B is(A) (B) (C) 2 (D) 4

69. B

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70. Number of lines that are parallel to 2x + 6y - 7 = 0 and have an intercept of 10 units between the co-ordinate axes is(A) 1 (B) 2(C) 4 (D) Infinite

70. B

71. The co-ordinates of point P on the line y = x for which PA + PB is minimum where A = (1, 3), B (3, 2) is (A) (2, 2) (B) (1, 2)

(C) (D)

71. D

72. The equation of the circle which touches the axes of co-ordinates and the line + = 1 and whose centre

lies in the first quadrant is x2 + y2 – 2cx – 2cy + c2 = 0 where c is equal to(A) 4 (B) 2(C) 3 (D) 6

72. D

73. The locus of a point which is equidistant from (-a, 0) and x = a is(A) y2 = 4ax (B) y2 = -4ax(C) x2 + 4ay = 0 (D) x2 - 4ay = 0

73. B

74. The equation of the line which represents the mirror image of the line 2x – 3y + 6 = 0 in y = – 1 is(A) 2x – 3y + 12 = 0 (B) 2x + 3y – 12 = 0(C) 2x + 3y + 12 = 0 (D) None of these

74. (C)BC = CDÞ D º (0, – 4)

Equation of AD: y + 4 = (x – 0)

Þ 2x + 3y + 12 = 0

75. The orthocentre of the triangle formed by the lines 4x + 5y – 12 = 0, 2x – y + 7 = 0 and 5x – 4y + 10 = 0 is at(A) (2/21, 100/21) (B) (– 2/21, 100/21)(C) (2/41, 100/41) (D) (– 2/41, 100/41)

75. (D)The given triangle is right angled because 4x + 5y – 12 = 0 and 5x – 4y + 10 = 0 are perpendicular.Þ Orthocentre is the point of intersection of these two lines.Þ (– 2/41, 100/41) is the orthocentre.

76. The co-ordinates of middle points of the sides of a triangle are (4, 2) (3, 3) and (2, 2). Then the co-ordinates of its centroid are

(A) (B) (3, 3)

(C) (4, 9/2) (D) none of these

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76. A

Centroid = = .

77. The tangents are drawn from the points of a straight line 3x + 4y = 24 to the curve

x2 + . Then all the chords of contact passes through a fixed point which lies on

(A) 16 x – 3y = 0 (B) 9y2 = 32 x(C) 24 x + 24 y = 19 (D) none of these .

77. (A), (B), (C)

Let the coordinates of a point lying on the straight line 3x + 4y = 24 is

Equation of the chord of contact is tx + (24 – 3t) = 1

Þ (24 y – 16) + t(16 x – 3y) = 0Þ this line always passes through the fixed point which is the point of intersection of the lines 24y – 16 = 0 and 16x – 3y = 0

fixed point º ,

which lies on 16x – 3y = 0, 9y2 = 32 x and 24 x + 24 y = 19.

78. Number of straight lines passing through (4, -5) and having distance from (-2, 3) equal to 12 is______________.

78. Straight line through (4, -5) be y + 5 = m (x –4)-mx + y + 5 + 4m = 0

Then

(6m + 8)2 = 144 (1 + m2)36m2 + 64 + 96m = 144 + 144m2

108m2 – 96m + 80 = 0 27m2 – 24m + 20 = 0Discriminant of the equation D = (24)2 –4 ´ 20 ´ 27 < 0It has no solution.

79. The slope of the line which belongs to family of lines (1 + l)x + (l – 1)y + 2(1 – l) = 0 and makes shortest

intercept on x2 = 4y – 4, is

(A) (B) 1

(C) 0 (D) 2

79. CFamily of line passes through focus hence latus rectum will make shortest intercept.

80. The equation of pair of straight lines perpendicular to 2y2 – 3xy – 2x2 + 3x + y – 1 = 0 and intersecting at (1, 0) is (A) 2y2 – 2x2 + x + 2y + 1 = 0 (B) 2y2 – 2x2 – 3xy + 4x + 3y – 2 = 0(C) x2 – y2 + 3xy + x – y + 2 = 0 (D) none of these

80. BLines parallel to given pair of straight line and passing through origin 2y2 – 3xy – 2y2 = 0 Þ (y – 2x)(2y + x) = 0.Hence pair of straight lines (y – 2(x – 1))(2y + x – 1) = 0.

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81. If the equation 2x + 3y + 1 = 0, 3x + y – 2 = 0 and ax + 2y – b = 0 are consistent then roots of the equation ax2 – bx – 2 = 0 is/are(A) 1 (B) –1(C) ± 1 (D) none of these

81. A If equations are consistent then

= 0 Þ b – a + 2 = 0.

Hence x = 1 is one of the root.

82. If (sin q, cos q); (q Î [0, 2p]) and (1, 4) lie on the same side of or on the line x – y + 1 = 0, then maximum value of sin q will be(A) 1 (B) 0

(C) (D) 1 –

82. Bsince .1 – 4 + 1 < 0, so

sin q – cos q + 1 £ 0

Þ sin q – cos q £ – Þ sin £ –

Þ Þ £ q £ 2p

Þ maximum value of sin q is 0.

83. If the quadrilateral Q1 formed by joining mid points of sides of another quadrilateral Q2 is cyclic, then Q1 is necessarily a(A) square (B) rectangle(C) rhombus (D) none of these

83. BWe know that quadrilateral Q1 is a parallelogram and Q1 is also cyclic so it becomes a rectangle.

84. Q and P are two points in I and IV quadrants respectively such that OQ is perpendicular to OP and OQ = 2OP (where O is origin). If circles on OP and OQ as diameter intersect at D and slope of line OP is tan a, then the slope of the line OD is

(A) (B)

(C) (D)

84. A

Since ÐODQ = ÐODP =

Þ Q, D, P are collinear Þ slope of line OD

= –

=

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85. In a triangle ABC, the bisectors of angles B and C lie along the lines x = y and y = 0. If A is (1, 2), then the equation of line BC is (A) 2x + y = 1 (B) 3x – y = 5(C) x – 2y = 3 (D) x + 3y = 1

85. BReflections of A in the two angle bisectors will on the line BC, so (2, 1) and (1, –2) will lie on BC.

Equation of BC will be y + 2 = i.e. 3x – y = 5.

86. If the vertices of a triangle are ( , 0), ( , ) and (2, 1) then the orthocentre of the triangle is (A) ( , 0) (B) (0, 0)(C) ( , ) (D) none of these

86. CThe circumcentre of the triangle is (0, 0) as all the vertices lie on the circle x2 + y2= 5.So the orthocentre will be (sum of x coordinates, sum of y coordinates).

87. One of the diagonals of a square is the portion of the line intercepted between the axes, then the

extremities of the other diagonal are (A) (5, 5), (-1, 1) (B) (0, 0), (4, 6)(C) (0, 0), (-1, 1) (D) (5, 5), (4, 6)

87. AExtremities of the given diagonal are (4, 0) and (0, 6)

Þ slope of this diagonal = –

Þ slope of other diagonal =

Þ equation of the other diagonal is

for the extremities of the diagonal r = Þ x – 2 = ± 3, y – 3 = ± 2Þ x = 5, –1 and y = 5, 1Þ the extremities of the diagonal are (5, 5), (–1, 1).

88. The coordinates of the point where, if the origin be shifted, the equation y2 + 4y + 8x - 2 = 0 will not contain linear term in y and the constant term are

(A) (1, -2) (B)

(C) (D) (1, 1)

88. CLet the origin be shifted to (a, b). So the given equation becomes (y + b)2 + 4(y + b) + 8(x + a) – 2 = 0

so, 2b + 4 = 0 and b2 + 4b + 8a – 2 = 0 Þ (a, b) is .

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89. Two sides of a triangle are along the coordinates axes and the medians through the vertices (other than the origin) are mutually perpendicular. The number of such triangles is/are(A) zero (B) two(C) four (D) infinite

89. ANo such triangle is possible as the medians through the vertices of a right angled triangle (other than right angle) can not be perpendicular to each other.

90. The set of values of m for which it is possible to draw the chord y = x + 1 to the curve x2 + 2xy + (2 + sin2

a)y2 = 1, which subtends a right angle at the origin for some value of a is (A) [2, 3] (B) [0, 1](C) [1, 3] (D) none of these

90. ACombined equation of pair of lines through the origin joining the points of intersection of line y = x + 1 with the given curve is x2 + 2xy + (2 + sin a)y2 – (y – mx)2 = 0for the chord to subtend a right angle at the origin(1 – m) + (2 + sin2 a – 1) = 0 (as sum of the coefficients of x2 + y2 = 0)Þ sin2 a = m – 2Þ 0 £ m – 2 £ 1Þ 2 £ m £ 3.

91. The point on the y-axis at which the line segment joining two fixed points A (a, 0) and B (b, 0) subtends a maximum angle is (0, c), then a, c, b (a > 0, b > 0) are in(A) A.P (B) G.P(C) H.P (D) none of these

91. BAs per the given condition the circle through A and B should touch the y-axis at (0, c), then OC2 = OA . OB (O is origin)Þ c2 = ab.a, b, c are in G.P.

92. The plane base of a solid in the x-y plane is made up of the points (x, y) such that 0 £ x £ 1 and 0 £ y £ 1. The cross section of the solid perpendicular to the x-axis is always a square. Then the volume of the solid is

(A) 1 (B)

(C) (D)

92. AHere the solid is in the form of a cube of side length 1.Volume = 1 cubic unit.

93. The reflection of the point (t – 1, 2t + 2) in a line is (2t + 1, t). Then the equation of the line can be (A) x = y + 1 (B) x = y – 1(C) x = 2y + 1 (D) x = 2y – 1

93. BLet the line be y = mx + c.

Slope of line segment joining both the given points =

Þ m = 1.

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Also, lies on it.

Þ Þ c = 1.

94. If y2 + 3xy + 2x2 – 5y – 7x + 6 = 0 represents pair of straight lines then the point of intersection is (A) (-1, -1) (B) (1, -1)(C) (-1, 1) (D) none of these

94. DPut x =1 and y =1 then verify Therefore (1, 1) is point of intersection

95. The equation of the line bisecting the obtuse angle between y –x = 2 and y + x = 5 is

(A) (B)

(C) (D) none of these

95. A

Bisector of given lines are

Þ - ( +1) x + . . . . (1)and . . . . (2)slope of line y – x = 2 is ( m1 =1)

and slope of first bisector

Now tanq = > 1.

Therefore q > 45°

So the bisctor of obtuse angle is .

96. If the curves ax2 + 4xy + 2y2 + x + y + 5 = 0 and ax2 + 6xy + 5y2 + 2x + 3y + 8 = 0 intersect at four concyclic points then the value of a is (A) 4 (B) – 4(C) 6 (D) – 6

96. BAny second degree curve passing through the intersection of the given curves is ax2 + 4xy + 2y2 + x + y + 5 + l ( ax2 + 6xy + 5y2 +2 x + 3y + 8 ) = 0If it is a circle, then coefficient of x2 = coefficient of y2 and coefficient of xy = 0 a(1+ l) = 2 + 5l and 4 + 6 l = 0

Þ a = and l =

Þ a = = – 4.

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97. If lines L1 and L2 belong to family of lines x(2 + l) + y(3 + 2l) – (8 + 5l) = 0 and are at maximum distance from points (-1, -3) and (6, 1) respectively then angle between L1 and L2 is

(A) (B)

(C) (D) none of these

97. CFor given family of lines 2x + 3y – 8 + l(x + 2y – 5) = 0, the point of concurrency is P(1, 2)let A º (-1, -3), B º (6, 1)

slope of , slope of BP =

If angle between L1 and L2 be q then

Þ

98. In a triangle ABC, AB is parallel to y-axis, BC is parallel to x-axis, centroid is at (2, 1). If median through A is x – y = 1 then median through C is (A) x – 3y + 1 = 0 (B) 2x – 3y – 1 = 0 (C) x – 4y + 2 = 0 (D) 3x + 2y = 8

98. C

Here

Slope of AM =

………..(1)

Slope of CN =

So equation of median through C is x – 4y + 2 = 0Note: If perpendicular sides of right angled D is parallel to co-ordinates axes then ratio of slope of medians

through A and C = or 4 where ÐB = 90o.

99. Let P is any point in 1st quadrant on the ellipse , tangent is drawn at P which intersects co-ordinate

axes at A and B. If circle drawn on AB as diameter has minimum area then co-ordinate of point P is

(A) (B)

(C) (D) none of these

99. C

The equation of tangent at P (2cos q, sin q) is

Þ AB2 = (2 sec q)2 + (cosec q)2 = 4 + 1 + 4 tan2 q + cot2 q ³ 5 + 4 = 9 Þ AB ³ 3 For area of circle to be minimum AB = 3

22

Þ 4 tan2 q = cot2 q

Þ point

100. If be a point on whose focii are then perimeter of triangle is(A) 10 (B) 8(C) 18 (D) 25

100. (C)

Here

Perimeter of triangle .

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Group-II

1. Triangle ABC with a = 3, b = 4 and c = 5 slides on the coordinate axis with A and B on the positive x-axis and the positive y-axis respectively. The locus of vertex C is (A) 4x – 3y = 0 (B) 3x– 4y = 0(C) x – y = 0 (D) none of these

1. AOACB is a cyclic quadrilateral Þ ÐBOC = ÐBAC

Þ tan q =

Þ locus of C is 4x – 3y = 0.

2. A line passes through (1, 0). The slope of the line, for which its intercept between y = x – 2 and y = -x + 2 subtends a right angle at the origin, is (A) ± 2/3, (B) ± 3/2(C) ± 1 (D) none of these

2. DHomogenizing (y – x + 2) (y + x – 2) = 0 with y = mx – m we get

angle subtended at the origin is 90o \ coefficient of x2 + coefficient of y2 = 0 Þ m2 – 4 – m2 = 0 Þ 4 = 0 which is absurd hence ‘m’ is not a finite number Þ line is a vertical line.

3. If L1, L2 and L3 are three non-concurrent and non-parallel lines then the maximum number of points which are equidistant from all the three lines are (A) 1, (B) 2(C) 3, (D) 4

3. DL1, L2, L3 are 3 non-concurrent lines

\ A triangle will be formed by the lines \ Incentre and 3- ex-centres are the points which are equidistant from L1, L2 and L3.Hence (D) is the correct answer.

4. If distance of any point (x, y) from the origin is defined as d = maximum {|x|, |y|}, then the distance of the common point for the family of lines x(1 + l) + ly + 2 + l = 0 (l - parameter) from origin is(A) 1, (B) 2(C) (D) 0

4. BCommon point for the family of lines is (-2, 1) \ Its distance from the origin is maximum {|2|, 1} = 2Hence (B) is the correct answer.

5. The point R on the line passing through Q (- 3, 6) such that the triangle PQR is equilateral with P (1, 3), is

24

(A) (B)

(C) (D) none of these

5. AIn triangle PQR, PQ = 5 = RQ = PR

Let slope of QR is M

slope of PQ =

Þ = ±

Þ 4 m + 3 = (4 - 3 m) … (1)and 4 m + 3 = - (4 - 3 m) … (2)

From (1), m = tan q =

cos q = , sin q =

co-ordinates of R are given by = 5

R º

Same way for R¢, m = tan q = Þ cos q = , sin q =

Co-ordinates of R¢ given by = - 5

Þ R¢ º

6. The number of integral points satisfying all the inequalities simultaneously 2 x - y £ 4, x + 2 y - 4 £ 0 and 5 y - 15 x + 32 £ 0, are (A) 1 (B) 0(C) 2 (D) none of these

6. B2 y + x £ 4 … (1)2 x - y £ 4 … (2) 5 y - 15 x + 32 £ 0… (3) 2 y + x = 4, 2 x - y = 4, 5 y - 15 x + 32 = 0 are three concurrent lines. So the only point satisfying all the inequalities is the point of

concurrency. Which is

So, number of integral points = 0.

7. Find the number of triangles formed by the lines represented by x3 - x2 - x- 2 = 0 and x y2 + 2 x y + 4 x - 2 y2 - 4 y - 8 = 0 (A) one (B) two(C) three (D) none of these

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7. Dx3 - x2 - x- 2 = 0 Þ (x - 2) (x2 + x + 1) = 0 Þ x = 2 … (1) x y2 + 2 x y + 4 x - 2 y2 - 4 y - 8 = 0 Þ (x - 2) (y2 + 2 y + 4) = 0 Þ x = 2 … (2) Both the equation represent same line. So number of triangle formed are zero.

8. Distance of origin from line (1 + )y + (1 – )x = 10 along the line y = x + k is

(A) (B) 5 + k

(C) 10 (D) 5

8. DAngle between both line is 450

OP = OP¢ = ´ = 5

9. If there is possible to draw a line which belong to all the given family of lines y – 2x + 1 + l1(2y – x – 1) = 0, 3y – x – 6 + l2(y – 3x + 6) = 0, ax + y – 2 + l3(6x + ay – a) = 0, then (A) a = 4 (B) a = 3(C) a = –2 (D) a = 2

9. AFirst two family of lines passes through (1, 1) and (3, 3) respectively Þ point of intersection of lines belonging to third family of lines will lie on line y = x Þ ax + x – 2 = 0 and 6x + ax – a = 0

or Þ a2 – a – 12 = 0 (a – 4) (a + 3) = 0

10. If = 1 and = 1 intersect the axes at four concylic points and a + b = c + d, then these lines always

intersect at (a , b, c, d > 0) (A) (1, 1) (B) (1, - 1) (C) (- 1, 1) (D) (- 1, - 1)

10. AFor points to be concyclic ac = bd also a + b = c + d

Þ = by compedendo

Þ b = c

similarly a = d Þ lines intersect at (1, 1)

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