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- 37Unit-11 Lines1. The equation of line equidistant from the points A(1, –2) and B(3,4) and making congruentangles with the coordinate axes is . . .(a) x +y = 1 (b) y – x + l = 0 (c)y – x – 1 = 0 (d) y – x = 22. The equation of line passing through the point (–5,4) and making the intercept of length25between the lines x + 2y – 1 = 0 and x + 2y + 1 = 0 is . . .(a) 2x – y + 4 = 0 (b) 2x – y –14 = 0 (c) 2x – y + 14 = 0 (d) None of these3. The equation of line containing the angle bisector of the lines 3x – 4y – 2 = 0 and 5x –12y + 2 = 0 is . . .(a) 7x + 4y – 18 = 0 (b) 4x – 7y – 1 = 0 (c) 4x – 7y + 1 = 0 (d) None of these4. The equation of line passing through the point of intersection of the lines3x – 2y = 0 and 5x + y – 2 = 0 and making the angle of measure tan–1(–5) with the positivedirection of x – axis is . . .(a) 3x – 2y = 0 (b) 5x + y – 2 = 0 (c) 5x + y = 0 (d) 3x + 2y + 1 = 05. If for a + b + c 0, the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 areconcurrent, then . . .(a) ab + be + ca = 0 (b)b c+ + 1b caa(c) a = b (d) a = b = c6. The equation of line passing through the point (1,2) and making the intercept of length 3units between the lines 3x + 4y = 24 and 3x + 4y = 12, is . . .(a) 7x – 24y + 41 = 0 (b) 7x + 24y = 55 (c) 24x – 7y = 10 (d) 24x + 7y – 38 = 07. If (a, a2) lies inside the angle between the lines y =2x, x > 0 and y = 3x, x > 0,then a . . .(a) 123, (b) (3,) (c) 12, 3(d) 120,8. If P(–1,0), Q(0,0) andR(3, 3 3), then the equation of bisector of PQR is . . .(a)32y 0x (b)32y 0x (c)3 y 0x (d)3y 0x
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- 389. If the non zero numbers a, b,c are in harmonic progression, then the liney 10b cxa passes through the point . . .(a) (1,–2) (b) (–1,–2) (c) (–1,2) (d)121,10. A line passing through 0(0,0) intersect the parallel lines 4x + 2y = 0 and 2x + y + 6 = 0 atP and Q respectively, then in what ratio does 0 dividePQfrom P ?(a) 1 : 2 (b) 3 : 4 (c) 2 : 1 (d) 4 : 311. The points on the line 3x – 2y – 2 = 0, which are 3 units away from the line3x + 4y – 8 = 0 are . . .(a) 13(3, 3), 3, (b) 7 312 3 23, , , (c) 712 3,3 , ,3(d) (3,1),(1,3)12. If A(1, –2), 5(–8,3), A–P–B and 3 AP = 7AB, then P = . . .(a) 41322, (b) 41322,(c) not possible (d) None of these13. For the collinear points P – A – B, AP = 4AB, then P dividesABfrom A in the ratio.....(a) 4 : 5 (b) – 4 : 5 (c) –5 : 4 (d) –1 : 414. If the length of perpendicular drawn from (5,0) on kx + 4y = 20 is 1, then k = . . .(a)1633, (b)1633, (c)1633, (d)1633, 15. If the lengths of perpendicular drawn from the origin to the lines xcos– ysin=sin2aand xsin+ ycos= cos2are p and q respectively, then p2+ q2= . . .(a) 4 (b) 3 (c) 2 (d) 116. The points onY – axis at a distance 4 units from the line x + 4y = 12 are . . .(a) (3 14, 0) (3 14, 0) (b)( 3 17, 0) (3 17, 0) (c)(0, 3 17)(d) (0, 3 17) (0, 3 17) 17. A base of a triangle is along the line x = b and its length is 2b. If the area of triangle is b2,then the vertex of a triangle lies on the line . . .(a) x =–b (b) x = 0 (c) x =b2(d )x = b18. Shifting origin at which point the transformed form of x2+ y2– 4x – 8y – 85 = 0 wouldbe x2+ y2= k?(a) (2,4) (b) (–2, –4) (c) (2, –4) (d) (–2,4)
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- 3919. A(1,0) and B(–1,0), then the locus of points satisfying AQ – BQ = ±1 is . . .(a) 12x2+ 4y2= 3 (b) 12x2– 4y2= 3 (c) 12x2– 4y2= –3 (d) 12x2+ 4y2= –320. A rod having length 2c moves along two perpendicular lines, then the locus of the mid pointof the rod is . . .(a) x2– y2= c2(b) x2+ y2= c2(c) x2+ y2= 2c2(d) None of these21. Consider a square PQR having the length of side a, where O(0,0). The sidesOPandORare along the positive X – axis and Y – axis respectively. If A and B are the mid pointsofPQandQRrespectively, then the angle betweenOAandOBwould be... . .(a) cos–135(b) tan–1 43(c) cos–134(d) sin–13522.3 y 2x is the equation of line containing one of the sides of an equilateral triangleand if (0,–1) is one of the vertices, then the length of the side of the triangle is . . .(a)3(b)2 3(c)32(d)2323. If the point t t2 21 , 2 lies between the two parallel lines x + 2y = 1 and2x + 4y = 15, then the range of t is . . .(a)56 20 t (b)4 23t 0 (c)4 2 5 23 6t (d) None of these24. If two perpendicular lines passing through origin intersect the lineyb1, 0, b 0xaa atA and B, then1 12 2OA OB.......... (a)2 21 1ba(b)2 2bbaa (c)2 22 2bbaa(d) None of these25. The equation of a line at a distance5units from the origin and the ratio of the interceptson the axes is 1 : 2, is . . .(a) 2x + y + 5 = 0 (b) 2x + y + 5 = 0 (c) x – 2y + 5 = 0 (d) None of these26. For any values of p and q, the line (p + 2q)x + (p – 3q)y – p – q passes through whichfixed point ?(a) 3 52 2,(b) 2 25 5,(c) 3 35 5,(d) 325 5,
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- 4027. If A(x1,y1), B(x2,y2) and P(tx2+ (1 – t)x1, ty2+ (1 – t)y1) where t < 0, then P dividesABfrom A in the ratio . . .(a) 1 – t (b)t 1t(c)t1 t(d) t – 128. A(1,2), B(5,7) and P(x,y)AB, then y – x – 1 is . . .(a) < 0 (b) > 0 (c) = 0 (d) –329. A(2,3), B(4,7) and P(x,y)AB, then the maximum value of 3x + y is . . .(a) 19 (b) 9 (c) –19 (d) –930. A(– 2,5), 5(6,2), thenAB AB .......... (a) {(8t – 2, 5 – 3t / t < 0) (b) {(8t – 2, 5 – 3t) / 0 < t < 1}(c) {(8t – 2, 5 – 3t) / t R – [0, 1]} (d) {(8t – 2, 5 – 3t) / t > 1}31. The p – form of the line3y 4 0x is(a)π π26 6xcos ysin (b)π π23 3xcos ysin (c)π π23 3xcos ysin (d)π π26 6xcos ysin 32. The length of side of an equilateral triangle is a. There is circle inscribed in a triangle.What is the area of a square inscribed in a circle ?(a)23a(b)26a(c)32a(d)22a33. If the lines x + 2ay + a – 0, x + 3by + 3 = 0 and x + 4cy + c = 0 are concurrent, then a,b, c are in . . .(a) A.P. (b) H.P. (c) G.P. (d) A.G.P34. The foot of perpendicular drawn from (2,3) to the line 4x – 5y – 34 = 0 is . . .(a) (6,–2) (b) 246 8241 41,(c) (–6,2) (d) None of these35. The equation of a line passing through (4,3) and the sum of whose intercepts is –1 is.....(a)y y2 3 2 11, 1x x (b)y y2 3 2 11, 1x x (c)y y2 3 2 11, 1x x (d)y y2 3 2 11, 1x x
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- 4136. A line intersects X – axis and Y – axis at A and B respectively. If AB = 15 and ABmakes atriangle of area 54 units with coordinate axes, then the equation ofABis . . .(a) 4x ± 3y = 36 or 3x ± 4y = 36 (b) 4x ± 3y = 24 or 3x ± 4y = 24(c) –4x ± 3y = 24 or – 3x ± 4y – 24 (d) –4x ± 3y = 12 or – 3x ± 4y – 1237. The angle between the lines xcos85° + ysin85° = 1 and xcos40° + ysin40° = 2 is :(a) 90° (b) 80° (c) 125° (d) 45°38. If a1, a2, a3and b1, b2, b3are in geometric progression and their common ratios are equal,then the pointsA(a1, b1), B(a2,b2) and C(a3,b3) . . .(a) lie on the same line (b) lie on a circle (c) lie on an ellipse (d) None of these39. The image of the point (4, –13) in the line 5x + y + 6 = 0 is . . .(a) (1,2) (b) (3,4) (c) (–4,13) (d) (–1, –14)40. If the lines x + (a – l)y + 1 = 0 and 2x + a2y –1 = 0 are perpendicular then . . .(a) | a | = 2 (b) 0 < a < 1 (c) –1 < a < 1 (d) a = –141. If x + 3y – 4 = 0 and 6x – 2y – 7 = 0 are the lines containing the diagonals of aparallelogram PQRS, then parallelogram PQRS is . . .(a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus42. For a + b + c = 0, the line 3ax + 4by + c = 0 passes through the fixed point . . .(a) 1 13 4, (b) 1 13 4, (c) 1 13 4,(d) 1 13 4, 43. If 3l + 2m + 6n = 0, then the family of lines lx + my + n = 0 passes through the fixedpoint . . .(a) (2,3) (b) (3,2) (c) 1 12 3,(d) 1 13 2,44. If the lines x + y + r = 0 andx – 5y = 5 are identical then+ r = . . ,(a) –4 (b) 4 (c) 1 (d) –145. If the x – coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1is an integer, then the integer value of m is . . .(a) 2 (b)0 (c) 4 (d) 146. If (4,5) is the foot of perpendicular on the line l, then the equation of the line l wouldbe . . .(a) 4x + 5y + 41 = 0 (b) 4x – 5y + 9 = 0 (c) 4x + 5y – 41 = 0 (d) None of these47. The y – intercept of the line y + y1= m(x – x1) is . . .(a) – (y1+ mx1) (b) y1– mx1(c)1 1y mmx(d) None of these
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- 4248. The locus of mid points of the segment intercepted between the axes by the linexseca + ytana = p is . . .(a)2 22 2p p4 4y1x (b)222 2yp p4x (c)2 22 21p px y (d)2 22 21p p4x 4y 49. If the y – intercept of the perpendicular bisector of the segment obtained by joiningP(l,4) and Q(k, 3) is –4 then k = . . .(a) 1 (b) 2 (c) –2 (d) –450. The y – intercept of the line passing through the point (2,2) and perpendicular to the line3x + y – 3 = 0 is . . .(a)34(b)43(c)43 (d)3451. The line parallel to the X – axis and passing through the intersection of the linesax + 2by + 3b = 0 and bx – 2ay – 3a = 0 where (a, b) (0,0) is :(A) above the X – axis at a distance of23from it(B) above the X – axis at a distance of32from it(C) below the X – axis at a distance of23from it(D) below the X – axis at a distance of32from it52. A square of side a lies above the x – axis and has one vertex at the origin. The side passingthrough the origin makes an angle a with the positive direction of x – axis.The eq. of its diagonal not passing through the origin is :(A) y(cos+ sin) + x(sin– cos) = a (B) y(cos+ sin) + x(sin+ cos) = a(C) y(cos+ sin) + x(cos– sin) = a (D) y(cos– sin) – x(sin– cos) = a53. If P and Q dividesABfrom A in the ratiosand –, then A dividesPQfrom p in theratio . . . . . . .(a)1+1(b)11(c)22(d)2254. The nearest point on the line x – 3y + 25 = 0 from the origin is . . .(a) (–4,5) (b) (–4,3) (c) (4,3) (d) None of these55. If the slope of a curve is constant, then the graph of a curve in the plane is . . .(a) line (b) parabola (c) hyperbola (d) none of these56. If 5x + 12y + 13 = 0 is transformed into xcos+ ysin= p, then= ?(a) cos–1 513(b) sin–1 1213(c) tan–1 125(d) tan–1 125
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- 4357.I fP(–1,0), Q(0,0) andR(3,3 3)are given points, then the equation of the bisector ofPQR is . . .(a)32yx (b)320x y (c)3 y 0x (d)3y 0x 58. For the line y – y1= m(x – xt), m and x1are fixed values, if different lines are drawnaccording to the different value of y1;then all such lines would be . . .(a) all lines intersect the line x = x1(b) all lines pass through one fixed point(c) all lines are parallel to the line y = x1(d) all lines will be the set of perpendicular lines59. If the length of perpendicular drawn from origin to a line is 10 and56 then theequation of line would be . . .(a)3 yx (b)3 yx (c)3 y 20 0x (d)3 y 20 0x 60. Find the equation of line making a triangle of area503units with two axes and on which aperpendicular from origin makes an anglewith positive direction of x – axis.(a)3yx (b) x – y = 10 (c)3 y 5x (d)3 y 10x 61. If 2x + 2y – 5 = 0 is the equation of the line containing one of the sides of an equilateraltriangle and (1,2) is one vertex, then find the equations of the lines containing the othertwo sides.(a)y (2 3) 3, y 3) 3x x (b)y (2 3) 3, y 3) 3x x (c)y (2 3) 3, y 3) 3x x (d)y (2 3) 3, y 3) 3x x 62. Find the equation of line passing through the point ( 3, 1) and at a distance2unitsfrom the origin.(a)( 3 1) ( 3 1) or ( 3 1) ( 3 1)yx y x (b)( 3 1) ( 3 1) or ( 3 1) ( 3 1)yx y x (c)( 3 1) ( 3 1) or ( 3 1) ( 3 1)yx y x (d)( 3 1) ( 3 1) or ( 3 1) ( 3 1)yx y x
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- 4463. If (3,–2) and (–2,3) are two vertices and (6,–1) is the orthocentre of a triangle, then thethird vertex would be . . .(a) (1,6) (b) (–1,6) (c) (1, –6) (d) none of these64. The circumcentre of the triangle formed by the lines x + y = 0, x – y = 0 and x – 7 = 0 is . .(a) (7,0) (b) (3.5,0) (c) (0,7) (d) (3.5,3.5)65. If1 1 1b c, ,aare in arithmetic sequence, then the liney1b c0xa passes through the fixedpoint . . .(a) (–1,–2) (b) (–1,2) (c) 121, (d) (1,–2)66. Find the slope of the line passing through the point (1,2) and the point of intersection ofthis line with the line x + y + 3 = 0 is at a distance3 2units from (1,2).(a)13(b) 1 (c)3(d)3 1267. The angle between the lines x = 3 and3 5x y is . . .(a)(b)3(c)4(d)268. The angle between the lines y = e and3 y 5x is . . .(a)(b)5(c)(d)369.T he angl e betw een the lines{(x, 0)/xR} and {(0,y)/ yR} is . . .(a)2(b)2(c) 0 (d) 70. If the point t t2 21 , 2 lies between the two parallel lines x + 2y = 1 and 2x + 4y = 15,then the range of t is . . .(a)56 20 t (b)4 23t 0 (c)4 2 5 23 6t (d) None of these71. If 2x + 3y = 8 is perpendicular to the line (x + y + 1) +(2x – y – 1) = 0, then= ?(a) –5 (b)32(c) 5 (d) 072. If the line (a + l)x + (a2— a — 2)y + a = 0 is parallel to Y – axis, then a = . . .(a) –1 (b) 2 (c) 3 (d) 1
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- 4573. The equation of a straight line passing through the point (–5, 4) and which cut off anintercept of2unit between the lines x + y + 1 = 0 and x + y – 1 = 0 is(a) x – 2y + 13 = 0 (b) 2x – y + 14 = 0 (c) x – y + 9 = 0 (d) x – y + 10 = 074. If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS then(a) a = 2, b = 4 (b) a = 3, b = 4 (c) a = 2, b = 3 (d) a = 2, b = 575. The sum of squares of intercepts on the axes cut off by the tangents to the curve2 2 23 3 3yx a (a > 0) at8 8,a ais 2. Thus a has the value.(a) 1 (b) 2 (c) 4 (d) 876. If two vertices of a trinangle eare (5, –1) and (–2, 3) and if its orthocentre lies at the originthen the cooridnates of the third vertex are(a) (4, 7) (b) (–4, –7) (c) (2, –3) (d) (5, –1)77. Line ax + by + p = 0 makes anglewhich+, R .x cos ysin p p If these lines andthe line0x sin y cos are concurrent then(a) a2+ b2= 1 (b) a2+ b2= 2 (c) 2(a2+ b2) = 1 (d) a2– b2= 278. A straight line passess through a point A(1, 2) and makes an angle 600with the x–axis. Thisline intersect the line x + y = 6 at P. Then AP will be(a)3( 3 1)(b)3( 3 1)(c)( 3 1)(d)3 379. The image of origin in the line x + 4y = 1 is(a)8217 17,(b)8217 17, (c)8217 17, (d)8217 17,80. Orthocentre of triangle with vertices (0, 0), (3, 4) and (4, 0) is(a)543,(b) (3, 12) (c)343,(d) (3, 9)81. The equation of three sides of triangle are x = 2, y + 1 = 0 and x + 2y = 4. The coordinatesof the circumcentre of the triangle is(a) (4, 0) (b) (2, –1) (c) (0, 4) (d) (–1, 2)82. If a, b, c are in A.P. then ax + by + c = 0 represents(a) a single line (b) a family of concurrent lienes(c) a family of parallel lines (d) a family of circle83. A(4, 0), B(0, 3), C(6, 1) be vertices of triangle ABC. Slope of bisector of angle C will be(A)3 2 7(b)5 2 7(c)6 2 7(d) none
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- 4684. The locus of the variable point whose distance from (–2, 0) is23times its distance fromthe line92x is(a) ellipse (b) parabola (c) circle (d) hyperbola85. The line 3x – 4y + 7 = 0 is rotated through an anglein the clockwise direction aboutthe point (–1, 1). The equation of the line in its new position is(a) 7y + x – 6 = 0 (b) 7y – x – 6 = 0 (c) 7y + x + 6 = 0 (d) 7y – x + 6 = 086. The area of the triangle formed by the point (a, a2), (b, b2), (c, c2) is ..... (a, b, c are threeconsecutive odd integers)(a)12(a – b) (b – c) sq unit (b) 8 sq unit(c) 16 sq unit (d)12(a – b) (b – c) (a + b + c) sq unit87. The straight line 7x - 2y + 10 = 0 and 7x + 2y – 10 = 0 forms an isosceles triangle withthe line y = 2. Area of the triangle is equal to :(a)157sq unit (b)107sq unit (c)187sq unit (d)1013sq unit88. In triangle ABC, equation of right bisectors of the sidesABandACare x + y = 0 andy – x = 0 respectively. If A = (5, 7) then equation of side BC is(a) 7y = 5x (b) 5x = y (c) 5y = 7x (d) 5y = x89. The equations of the two lines each passing through (5, 6) and each making an acute angleof 450with the line 2x – y + 1 = 0 is(a) 3x + y – 21 = 0, x – 3y + 13 = 0 (b) 3x + y + 21 =0, x + 3y + 13 = 0(c) y = 2x, y = 3x (d) 3x + y – 21 = 0, x – 3y – 13 = 090. If the equation of base of an equilateral triangle is 2x – y = 1 and the vertex is (–1, 2),then the length of the side of the triangle is(a)203(b)215(c)815(d)15291. Four points (x1, y1), (x2, y2), (x3, y3) and (x4, y4) are such that 42 21i iix y 2(x1x3+ x2x4+ y1y2+ y3y4). Then these points are vertices of(a) parallellogram (b) Rectangle (c) Square (d) Rhombus
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