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Straight Line, , JEE (Main+Advanced), Exercise – I, , Section (A) : Coordinate system, Distance formula,, Section formula, 1., , , , , Find the distance between points P  2, , ,  , ,  6, , and Q  3,, (a) 7, , (b), , , , 6, , 7, , 4., , (c) 19, (d) 19, The coordinates of a point are (0,1) and the, ordinate of another point is -3. If the distance, between the two point is 5 then the abscissa of, another point is (a) -3, (b) 3, (c)  3, (d) 1, The three points ( 2, 2), (8, 2) and ( 4, 3), are the vertices of, (a) an isosceles triangle, (b) an equilateral triangle, (c) a right angled triangle, (d) None of these, The quadrilateral formed by the points, , 5., , (a) rectangle, (c) square, The points, , 6., , (a) square, (b) rectangle, (c) rhombus, (d) None of these, If (3, 4) and ( 6,5) are the extremities of the, , 2., , 3., , 7., , 8., , (a, b),(0, 0), (a, b) and (ab, b 2 ) is (b) parallelogram, (d) None of these, , A(4, 1), B (2, 4), C (4, 0) and, , D(2,3) are the vertices of, , diagonal of a parallelogram and ( 2,1) is its, third vertex, then its fourth vertex is (a) ( 1,0), (b) ( 1,1), (c) (0, 1), (d) (0,1), If x-axis divides the line joining (3,4) and (5,6), in the ratio  :1 then  is-, , 3, (a) , 2, 3, (c), 4, , 2, (b) , 3, 1, (d), 3, , 9., , The points which trisect the line segment joining, the points (0,0) and (9,12) are, (a) (3,4),(6,8), (b) (8,6),(0,2), , 10., , (c) (1,3)(2,5), (d) (4,0),(0,3), A point on the line joining points (0,4) and (2,0), dividing the line segment externally in ratio 3 :, 2, is (a) (3,-4), (b) (6,-8), , 3 8, 5 5, , (d)  , , , 1 , 2 , (c) (2, 3), ,  1,  2, (d) (1, 4), , P and Q are points on the line joining A(-2,5), and B (3,1) such that AP  PQ  QB. Then the, mid point of PQ., (a)  ,3 , , 11., , 12., , 8 3, 5 5, , (c)  , , , , , , (b)   , 4 , , The ratio in which the join of the points (1,2), and (-2,3) is divided by the line 3 x  4 y  7 is -, , (a) 4 :1, (b) 3: 2, (c) 3 :1, (d) 7 : 3, Find the harmonic conjugate of point R(2,4), with respect to the points P (2, 2) and Q (2,5)., (a) (4, 2), , (b) ( 2,3), , (c) (2,8), (d) (8, 2), Section (B): Area of triangle , Locus, Change of, origin, Slope of line, Co linearity, 13., The area of the triangle formed by the mid, points of sides of the triangle whose vertices are, (2,1), ( 2,3), (4, 3), is, (a) 1.5 sq. units, (b) 3 sq. units, (c) 6 sq. units, (d) 12 sq. units, 14., The vertices of a triangle ABC are, ( , 2  2 ), (  1, 2 ) and (4   , 6, 2 )., If its area be 10 units then number of integral, values of  is, (a) 1, (b) 2, (c) 3, (d) 4, 15., Line, segment, joining, (5,0), and, (10 cos  ,10sin  ) is divided by a point P in, ratio 2 : 3. If  varies then locus of P is -, , (a) ( x  3) 2  y 2  16, (b) ( x  3)2  y 2  16, (c) y 2  4 x, 16., , (d) y  5 x  12, A rod of length l slides with its ends on two, perpendicular lines. Find locus of its mid points., , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 1

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JEE (Main+Advanced), , l2, 4, 2, 2, 2, 2, 2, (c) 2 x  2 y  l (d) x  y  2l 2, (a) x 2  y 2  l 2, , 17., , 19., , 2, 2, (b) x  y , , Find the locus of a point which moves so that, sum of the squares of its distance from the axes, is equal to 3., (a) x 2  y 2  9, , 18., , (c) x  y  3, , (b) x 2  y 2  3, , (d) x 2  y 2  3, , At what point should the origin be shifted if the, coordinates of a points (9,5) become (-3,9), (a) (-12,4), (b) (-4,7), (c) (7,-4), (d) (12,-4), Find the new position of origin so that equation, , x  4 x  8 y  2  0 will not contain a term in x, 2, , and the constant term., , 3 , (b), 4 , 3, , (c)  2, , (d), 4, , If A(2,3), B(3,1) and, (a)  , 4 , , 20., , 21., , 3, ,  , 2 , 4, , 3, ,  2,  , 4, , C (5, 3) are three points,, , then the slope of the line passing through A and, bisecting BC is (a) 1/2, (b) -2, (c) -1/2, (d) 2, Slope of line joining points (5, 3) and, (c) 2  1, If the points, , 25., , 26., , (b) 1  2, , (d) 1  2, , (k , 2  2k ), (1  k , 2k ) and, , 1, 2, , 1, , 1, 2, , (a)  ,1, , (b), , (a) y  x sin , , (b) y  x tan , , (c) 1,2, (d) 1,3, Section (C) : Various forms of straight line, Point and, line, Angle between two lines, 23., The equation of a line passing through the origin, and the point (a cos  , a sin  ) is (c) y  x cos , (d) y  x cot , If the point (5,2) bisects the intercept of a line, between the axes, then its equation is -, , (c) 5 x  2 y  20 (d) 2 x  5 y  20, Equation to the straight line cutting off an, intercept unity from the positive direction of the, axis of y and inclined at 450 to the axis of X is (a) x  y  1  0 (b) x  y  1  0, (c) x  y  1  0, , 27., 28., , segment between the axes is bisected at this, point, then its equation is given by, , 30., , 31., , 32., , x y,  1, x1 y1, , (c) xy1  yx1  1, , (b), , x y,  2, x1 y1, , (d) 2( xy1  yx1 )  x1 y1, , Slope of line bisecting the angle between coordinate axes, is, (a) 3, (b) 2, (c) 1 (d) 5, Find equation of a straight line on which length, of perpendicular from the origin is four units and, the line makes an angle of 1200 with the positive, direction of x-axis., (a), , 29., , (d) x  y  2  0, , If a straight line passing through ( x1 , y1 ) and its, (a), , ( k  4, 6  2k ) are collinear, then possible, , values of k are, , 24., , (a) 5 x  2 y  20 (b) 2 x  5 y  20, , (k 2 , k  1) is ½, then k is, , (a) 1, , 22., , Straight Line, , 3x  y  0, , (b), , 3x  y  8, , (c) x  3 y  8 (d) x  3 y  8, The point (-4,5) is the vertex of a square and one, of its diagonals is 7 x  y  8  0 . The equation, of the other diagonals is, (a) 7 x  y  23  0 (b) 7 y  x  30, (c) 7 y  x  31, (d) x  7 y  30, The points A (1,3) and C (5,1) are the opposite, vertices of rectangle. The equation of line, passing through other two vertices and of, gradient 2, is, (a) 2 x  y  8  0 (b) 2 x  y  4  0, (c) 2 x  y  4  0 (d) 2 x  y  4  0, Equation of a straight line passing through the, origin and making with x-axis an angle twice the, size of the angle made by the line y  (0.2) x, with the x-axis, is :, (a) y  (0.4) x, (b) y  (5 / 12) x, (c) 6 y  5 x  0 (d) None of these, Equation of the line passing through the point, (1,-1) and perpendicular to the line 2 x  3 y  5, is, (a) 3 x  2 y  1  0 (b) 2 x  3 y  1  0, , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 2

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Straight Line, , JEE (Main+Advanced), (c) 3 x  2 y  3  0 (d) 3 x  2 y  5  0, , 33., , If the line passing through the points (4, 3) and, , 34., , then  is equal to (a) 4, (b) -4, (c) 1, (d) -1, The equation of a line parallel to 2 x  3 y  4, which makes with the axes a triangle of area 12, units, is (a) 3 x  2 y  12 (b) 2 x  3 y  12, , 35., , 36., , 37., , Which pair of points lie on the same side of, , 41., , The set of values of ‘b’ for which the origin and, the point (1,1) lie on the same side of the straight, , (2,  ) is perpendicular to the line y  2 x  3,, , (c) 2 x  3 y  6, , (d) 3 x  2 y  6, , Area of  formed by line 3 x  4 y  12  0 with, co-ordinate axis is, (a) 6, (b) 2, (c) 1, (d) 5, If the middle points of the sides BC, CA and AB, of the triangle ABC be (1,3), (5, 7) and ( 5, 7), respectively then the equation of the side AB is, (a) x  y  2  0 (b) x  y  12  0, , (c) x  y  12  0 (d) None of these, The distance of the point (2,3) from the line, 2 x  3 y  9  0 measured, along, a, line, , x  y  1  0 is :, , (a) 5 3, 38., , 40., , (c) 3 2, , (b) 4 2, (d) 2 2, , From (1,4) you travel 5 2 units by making 135, angle with positive x-axis (anticlockwise) and, then 4 units by making 1200 angle with positive, x-axis (clockwise) to reach Q. Find co-ordinates, of point Q., , line, a 2 x  aby  1  0a  R, b  0 are :, , (a) b  (2, 4), 42., , 43., , A straight line is drawn through the point (2,3), and is inclined at an angle of 30 with the xaxis. Find the coordinates of two points on it at a, distance 4 from P., , (d) (2  2 3,5)or (2  2 3,1), , The angle between the lines 2 x  3 y  5 and, , 3 x  2 y  7 is -, , (b) 300, , (c) 600, Given, , (d) 900, , points, A(4,5), B( 1, 4), C (1,3), D (5, 3) then, the, ratio of the segment into which AB is divided by, CD is, , 13, 20, 13, (d), 19, (b), , Section (D) : Centroid, Circumcenter Orthocenter,, Incenter, Excenter, 44., The line bx  ay  3ab cuts the coordinate axes, at A and B, then centroid of OAB is (a) (b, a ), (b) ( a, b), , 45., , 46., , (a) (2  2 3,5)or (2  2 3,1), (c) (2  2 3,5)or (2  2 3,1), , (d) (2, ), , 20, 13, 11, (c), 20, , 0, , , (b) (2  2 3,5)or (2  2 3,1), , (c) b  [0, 2], , (a), , (a) 6,9  2 3, , 39., , (b) b  (0, 2), , (a) 450, , 0, , ,  (b)  6,9  2 3 , (c)  6,9  2 3  (d)  6,9  2 3 , , 3x  8 y  7  0, (a) (0, 1)and (0, 0) (b) (4, 3)and (0,1), (c) ( 3, 4)and (1, 2) (d) ( 1, 1)and (3, 7), , 47., , (c) ( a / 3, b / 3), (d) (3a,3b), Circumcentre of a triangle whose vertex are, (0, 0), (4, 0) and (0,6) is -, , 4 , 3 , (c) (2, 3), (a)  , 2 , , (b) (0, 0), , (a) (2a, 2b), , (b)  , , , (d) (4, 6), The orthocentre of the triangle ABC is ‘B’ and, the circumcentre is S (a,b). If A is the origin,, then the co-ordinates of C are :, , (c), , , , a 2  b 2 ,0, , , , a b,  2 2, , (d) None, , The incentre of the triangle formed by, (0, 0), (5,12), (16,12) is, (a) (7, 9), , (b) (9, 7), , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 3

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JEE (Main+Advanced), , 48., , 49., , Straight Line, , (c) ( 9, 7), (d) ( 7, 9), If two vertices joining the hypotenuse of a right, angled triangle are (0,0) and (3,4), then the, length of the median through the vertex having, right angle is (a) 3, (b) 2, (c) 5/2, (d) 7/2, A variable straight line passes through a fixed, point (a,b) intersecting the co-ordinates axes at, A & B. If ‘O’ is the origin, then the locus of the, centroid of the triangle OAB is :, (a) bx  ay  3 xy  0, (b) bx  ay  2 xy  0, , 55., , (c) 3 x  y  15  0 (d) x  3 y  5  0, Section (F) : Angle bisectors, concurrent lines and, family of lines, 56., The equation of the bisector of the acute angle, between, the, lines, 3 x  4 y  7  0 and, , (c) ax  by  3 xy  0, , (d) ax  by  2 xy  0, Section (E) : Distance between parallel lines, Foot of, the perpendicular image of a point and Area of, parallelogram, 50., The, figure, formed, by, the, lines, , 57., , 2 x  5 y  4  0, 5 x  2 y  7  0, 2 x  5 y  3  0, and 5 x  2 y  6  0 is, , 51., , (a) Square, (b) Rectangle, (c) Rhombus, (d) None of these, Area of the parallelogram formed by the lines, y  mx, y  mx  1, y  nx and, y  nx  1, equals, (a), , 52., , 53., , 54., , mn, ( m  n) 2, , 1, (c), mn, , (b), , 2, mn, , 1, (d), mn, , 58., , 59., , The reflection of the point (4, 13) in the line, , 5 x  y  6  0 is, (a) (1, 14), (b) (3, 4), (c) (1, 2), (d) ( 4,13), The image of the point A(1, 2) by the line mirror, y  x is the point B and the image of B by the, line mirror y  0 is the point ( ,  ), then :, (a)   1,   2 (b)   0,   0, (c)   2,   1 (d) None of these, The foot of perpendicular drawn from point, (1,2) on the line L is (2,3), then equation of line, L is, (a) x  y  3  0 (b) x  y  5  0, , (c) x  y  5  0 (d) 2 x  y  5  0, A light beam emitting from the point A(3,10), reflects from the straight line 2 x  y  6  0 and, then passes through the point B(4,3). The, equation of the reflected beam is :, (a) 3 x  y  1  0 (b) x  3 y  13  0, , 60., , 61., , 12 x  5 y  2  0 is, (a) 11x  3 y  9  0 (b) 3 x  11 y  13  0, (c) 3 x  11 y  3  0 (d) 11x  3 y  2  0, , The equations of bisectors of two lines L1 & L2, are 2 x  16 y  5  0 and 64 x  8 y  35  0. If, the line L1 passes through (-11,4), the equation, of acute angle bisector of L1 & L2 is :, (a) 2 x  16 y  5  0, (b) 64 x  8 y  35  0, (c) data insufficient, (d) None of these, The lines ( p  q ) x  ( q  r ) y  ( r  p )  0,, , (q  r ) x  (r  p ) y  ( p  q )  0, (r  p ) x  ( p  q ) y  (q  r )  0 are, , (a) Parallel, (b) Perpendicular, (c) Concurrent, (d) None of these, The least positive value of t so that the lines, x  t  a, y  16  0 and y  ax are concurrent, is, (a) 2, (b) 4, (c) 16, (d) 8, The solution of equation of equations, x  y  10, 2 x  y  18, and 4 x  3 y  26, will, be, (a) Only one solution, (b) No solution, (c) Infinite solution, (d) None of these, The, lines, ax  by  c  0, where, , 3a  2b  4c  0, are concurrent at the point :, , 1 3, 2 4, , (a)  , , , (b) (1,3), , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 4

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JEE (Main+Advanced), , 62., , (c) (3,1), , Straight Line, , 3 1, 4 2, , (d)  , , , 69., , (b) ax  by  1  0, x  y  0, , parallel to the line 2 x  y  4 is -, , 63., , 64., , 65., , (c) 2 x  y  1  0 (d) None of these, The fix point through which the line, x(a  2b)  y (a  3b)  a  b always passes for, all values of a and b, is (a) (2,1), (b) (1, 2), , (c) (2, 1), (d) (1, 2), A straight line cuts intercepts from the, coordinate axes sum of whose reciprocal is 1/p., It passes through a fixed point (a) (1/ p, p ), (b) ( p,1/ p ), (c) (1/ p,1/ p ), (d) ( p, p), The line parallel to the x-axis and passing, through the intersection of the lines, ax  2by  3b  0 and bx  2ay  3a  0, where, , (a, b)  (0, 0) is, , 66., , (a) Above the x-axis at a distance of 3/2 form it, (b) Above the x-axis at a distance of 2/3 from it, (c) Below the x-axis at a distance of 3/2 from it, (d) Below the x-axis at a distance of 2/3 from it, Section (G) : Pairs of lines and homogenization, (b) x  y  0 & x  11 y  0, (c) x  y  0 & x  11 y  0, , 67., , 68., , 70., , (c) ax  by  1  0, (d) None of these, The, angle, , 71., , between, , the, , lines, , x  xy  6 y  7 x  31y  18  0 is, 2, , (a) 450, , 2, , (b) 600, , (c) 900, If, , (d) 300, the, , equation, , 2 x  kxy  3 y  x  4 y  1  0 represents, 2, , 2, , pair of lines, then the value of ‘k’ can be :, (a) 1, 5, (b) 3,5, 72., , 73., , The equation 4 x 2  24 xy  11y 2  0 represents, (a) x  y  0 & x  11 y  0, , equation ax 2  (a  b) xy  by 2  x  y  0 are, , (a) ax  by  1  0, x  y  0, , The equation of the line through the point of, intersection of the lines y  3 and x  y  0 and, (a) 2 x  y  9  0 (b) 2 x  y  9  0, , The equation of the lines represented by the, , 74., , (d) x  y  0 & x  11 y  0, If the slope of one line of the pair of the lines, , represented by ax 2  10 xy  y 2  0 is four times, the slope of the other line, then ‘a’ equals to, (a) 1, (b) 2, (c) 4, (d) 16, The combined equation of the bisectors of the, angle between the lines represented by, , (c) 1,3, If, , (d) 2,5, the, , equation, , 2 x  2 xy  y  6 x  6 y  c  0 represents, 2, , 2, , a, , a, , pair of lines, then ‘c’ is, (a) 2, (b) 3, (c) -3, (d) 1, The straight lines joining the origin to the points, of intersection of the line 2 x  y  1 and curve, , 3x 2  4 xy  4 x  1  0 include an angle :, , , (a), (b), 2, 3, , , (c), (d), 4, 6, , If distance between the pair of parallel lines, , x 2  2 xy  y 2  8ax  8ay  9a 2  0 is 25 2,, then ' a / 5' is equal to, (a) 4, (b) 2, (c) 3, (d) 1, , ( x 2  y 2 ) 3  4 xy is, , (a) y 2  x 2  0, , (b) xy  0, , (c) x 2  y 2  2 xy (d), , x 2  y 2 xy, , 2, 3, , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 5

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Straight Line, , JEE (Main+Advanced), , Answer Key, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, , b, c, c, d, b, a, b, a, b, a, a, c, a, a, a, b, b, d, c, c, , 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, , b, b, b, b, b, b, c, b, c, b, b, a, a, b, a, b, b, b, a, d, , 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, , b, d, b, b, c, a, a, c, a, c, d, a, c, b, b, a, a, c, d, a, , 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, , d, a, c, d, c, c, d, a, a, a, a, b, a, d, , ADDRESS : NEAR CANARA BANK, JAIL BYPASS ROAD, PADRI BAZAR, GORAKHPUR,, MOB: 6386566032,7992166350, , 6