Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y + = 0 Check: (-)+(-4) (-5) < 0 Eqn. to acute angle bisector x-4 y+7 -x-5 y+ = 5 x - y + 9 = 0 Q.No. The equation of the bisector of the angle between two lines x - 4y + = 0 and x - 5y + 7 = 0 which contains the points (-, 4) is: Option x + 7y - = 0 Option x - 7y + = 0 Option x + 7y + 9 = 0 Option 4 - x+4y- x-5 y+7 = 5 Eqn. to the bisector containing (0, 0) x-4 y+ x-5 y+7 = 5 Check: (-)-4 +) < 0 And (-)-5 4+7) < 0 (-, 4) lies is the opposite region as shown below: So, eqn. to the bisector containing (-, 4) is x-4 y+ x-5 y+7 = 5 x + 7y - = 0 Q.No. The line x + y - = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0. The equation of the other line is : Option x + y - = 0 Option x - y + = 0 Option 5x + 5y - = 0 Option 4 None Correct Answer Let the line is x + y - + λ(x - 7y + 5) = 0 Therefore, + λ - + - - -7λ = 7 -- + - λ + - -7 7 λ
λ = 4 Eqn. to line is 5x + 5y - = 0 Q.No. 4 Find the angle between there lines represented by the equation x - pxy + y = 0 Option sec - (p) Option tan - (p) Option cos - (p) x - p xy + y = 0 let the angle be tan θ tan θ= p - tan θ= p - sec θ= p sec θ= p θ=sec ( p) is one solution. - Q.No. 5 Option 4 The distance between the lines represented by the equation, x + xy+ y +4 x+4 y+=0 is: Option 4 Option Option 4 Correct Answer x + xy+ y +4 x+4 y+=0 Let the lines be x+ y+a =0 x+ y+a =0 A A = and A + A = 4 Distance between lines A =A = = L + ( ) = = (A +A ) -4A A Q.No. 6 If sum and product of the slopes of lines represented by 4x + hxy - 7y = 0 is equal to then h is equal to: Option -6 Option - Option -4 Option 4 4 Correct Answer 7y - hxy - 4x = 0
h -4 = h=- 7 7 Q.No. 7 If θis the angle between the lines represented by x - xy+ λy +x-5 y+=0, where λ is a real number, then cosec θ equals : Option 9 Option Option 0 Option 4 00 Correct Answer x - xy+ λy +x-5 y+=0 - -5 a=, h=, b= λ, g=, f =, c= For pair of straight lines : =0 45 5 9 9 λ + - - λ. -. =0 4 4 4 4 λ = 9-4 tan θ = = + cot θ= cosec θ=0 Q.No. 8 The difference of the slopes of the lines, x (sec θ -sin θ )-xy tan θ + y sin θ =0 Option - Option Option Option 4 Correct Answer y sin θ - xy tan θ + x (sec θ - sin θ ) = 0 tan θ sec θ - sin θ [ m - m]= -4. sin θ sin θ 4 = tan θ -(tan θ -sin θ ) sin θ sin θ = = sin θ is Q.No. 9 If pairs of straight lines, x - p xy - y = 0 and x - q xy - y = 0 be such that each pair bisects the angles between the other pair then : Option pq =- Option pq = - Option pq = - Option 4 p =- q
Correct Answer Eqn. to bisectors of x - pxy - y = 0 x -y xy = -p px + xy - py = 0 () is Equivalent to x - qxy - y = 0 () p = -q pq=- () Q.No. 0 Find the condition that the pair of straight lines joining the origin to the intersections of the line y = mx + c and the circle x + y = a may be at right angles. Option a = c ( + m ) Option a = c (- m ) Option c = a ( + m ) Option 4 c = a (- m ) Correct Answer y-mx Homogenising x + y = a with y= mx+ c x + y -a =0 c c x + c y - a y - a m x + amxy = 0 Sum of coeff. of x and y = 0 c + c -a - a m =0 a (+ m )=c Q.No. The vertices of OBCare respectively (0, 0), (-,-) and (-,-). The equation of line parallel to BC and at a distance form O which intersects OB and OC is: Option x+ y + =0 Option x- y + =0 Option x+y - =0 -(-H) Slope of OD= = (-+) Hence, Eq. of line x y + =- x+ y+ =0
Q.No. If the sum of the distance of a point from two perpendicular lines in a plane is then it locus is : Option Square Option Circle Option A straight line Option 4 Two intersecting line Let the two lines be ax + by + c = 0 - L bx - ay + c = 0 - L Let the point be h, k ah+ bk+ c bh+ ak+ c + = a + b a + b If the axes are along the line L and L then x + y = Clearly, the locus is square. Q.No. Given the points A(0,4) and B(0, -4), the equation of the locus of the point P(x, y) such that AP-BP =6 is : Option 9x - 7y + 6 = 0 Option 9x - 7y - 6 = 0 Option 7x - 9y + 6 = 0 Option 4 7x - 9y - 6 = 0 x +( y+4) - x +( y+4) =6 x +( y-4) + x +( y+4) - x +( y-4) x +( y+4) =6 ( x + y -) = x +( y-4) x +( y+4) -4x -4 y +4=x - y +56 6x -8 y +5=0 9x -7 y +6=0 Q.No. 4 A triangle ABC with vertices A(-, 0), B(-, /4) and C(-, -7/6) has its orthocentre H. Then the orthocentre of triangle BCH will be : Option (-,-) Option (, ) Option (-, ) Correct Answer 4
A will be orthocentre of HBC Q.No. 5 A variable straight line passes through a fixed point (a, b) intersecting the co-ordinates axes at A and B. If O is the origin then the locus of the centroid of the triangle OAB is : Option bx + ay - xy = 0 Option bx + ay - xy = 0 Option ax + by - xy = 0 Option 4 ax + by - xy = 0 Let (y- b) = m(x - a) be the variable line through (a, b). Let (h, K) be the centroid. b h= a- ; m K = b - am; ab = (a- h) (b- K) bx + ay = xy - required locus. Q.No. 6 Two vertices of a triangle are (4,-) and (-, 5). If the othocentre of the triangle is at (, ), then the coordinates of the third vertex are Option (, 0) Option (, 6) Option (-, 6) Option 4 (,-6) Correct Answer BE AC
y-5 -(-) =(-) x+ -4 x-5 y+ =0 ( ) Also CF AB gives x - y - 7 = 0 -() from ()&() (x, y) (, 6) Q.No. 7 Option Option Option Drawn from the origin are two mutually perpendicular straight lines forming an isosceles triangle together with the straight line, x + y = a. Then the area of the triangle is : a a 5 Option 4 none Correct Answer a - Let the slope of the lines bem&. m Then, + m- = m L+m - m m= or- Let ' θ ' be the equal angle - - =tan 0 θ =45 + a Length of perpendicular from (0,0) = 5
a a a Area of = = 5 5 5 Q.No. 8 The point (a, a + ) is a point in the angle between the lines x - y + = 0 and x + y - 5 = 0 containing the origin if : Option a ora - Option a (-,0), Option a (0,) Correct Answer (a -( a+)+) > 0and( a +( a+)-5) (-5) > 0 a(- a) > 0 & ( a+)( a-) < 0 a (-,0), Q.No. 9 Area of the quadrilateral formed by the lines x + y = is : Option 8 Option 6 Option 4 Option 4 none x + y = Locus of (x, y) is square Area of square = ( ) =8 Q.No. 0 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 : Option (a, b) Option a, b Option a + b,0 Option 4 None ( )
One of the vertices is orthocentre. So, it is right angle triangle at which is midpoint of AC. C ( a, b) Q.No. The equation of the line segment AB is y = x. If A and B lie on the same side of the line mirror x - y =, the image of AB has the equation Option x + y = Option 8x + y = 9 Option 7x - y = 6 Correct Answer Clearly, mirror could be bisector of angle between AB & A B. Let m be the slope of A B. m- - = L+m + m=7 or m= Image line passes through point of intersection of AB & mirror. ( y- x)+ λ (x-y-)=0-a'b' -(λ-) 6 slope=7= λ= L- λ 5 Hence image line: 7x - y = 6 Q.No. 4 A is a point on either of two lines y+ x =at a distance of units from their Point of intersection. The co-ordinates of the foot of perpendicular from A on the bisector of the angle between them are Option -, Option (0, 0) Option,
Option 4 (0, 4) Correct Answer Two lines are y+ x = y- x = Intersect at (0, ) Equations to bisectors y+ x- = y- x - x = 0 or y = 0 4 In PBA, B=90 &PA= 0 4 PB=PAcos60 = = B, There would be other point of same property on y = - B', by symmetry. For bisector x = 0 0 In POA, O=90 0 4 PO=PAcos0 = = Hence, such O points will have coordinates (0, 0) &(0, 4). Hence all options are correct. Q.No. On the portion of the straight line, x + y = 4 intercepted between the axes, a square is constructed on the side of the line away from the origin. Then the point of intersection of its diagonals has co-ordinates : Option (, ) Option (, ) Option (, ) Option 4 none Correct Answer
AB= 5 AD= 0 m+ Let m be slope of AD = m - m=,- Slope of AD = =tanθ To be coordinates of D 0+ 0., + 0 (,) 0 0 Q.No. 4 The equations of three lines, AB, CD and EF are, (b- c)x + (c- a)y + (a- b) = 0, (c- a) x + (a- b)y + (b- c) = 0 and (a- b)x + (b- c)y + (c- a) = 0. Which one of the following inferences is correct. Option The lines are parallel to each other Option AB and BC are perpendicular to EF Option All the lines are coincident Option 4 The lines are coincident Correct Answer 4 AB :(b- c)x + (c- a)y + (a- b) = 0 CD :(c- a)x + (a- b)y + (b- c) = 0 AB :(a- b)x + (b- c)y + (c- a) = 0 Since: AB + CD + EF = 0 Lines are coincident Q.No. 5 The base BC of a ABCis bisected at the point (p, q) and the equation to the side AB and AC are px + qy = and qx + py =. The equation of the median through A is : Option (p- q)x + (q- p)y + = 0 Option (p + q) (x + y) - = 0 Option (pq- ) (px + qy - ) = (p + q - )(qx + py -) Option 4 none Correct Answer Equation of median through A px + qy - + λ(qx + py - ) = 0 passes through (p, q) p + q - + λ(pq + pq - = 0) p + q - λ= - pq median : p + q - px+ qy-= ( qx+ py-)=0 -pq (pq- )(qx + py - ) = (p + q - )(qx + py -)
Q.No. 6 Given the family of lines, a (x + 4y + 6) + b(x + y + ) = 0. The line of the family situated at the greatest distance from the point P(, ) has equation : Option 4x + y + 8 = 0 Option 5x + y + 0 = 0 Option 5x + 8y + 0 = 0 Option 4 None a(x + 4y + 6) + b(x + y + ) = 0 always through intersection of (x + 4y + 6) = 0 and x + y + = 0 i.e. (-, 0). Most distant line would be perpendicular to PR. Eqn. to line y - = x + -0 + y + 4y + 8 = 0 Q.No. 7 Let P = (, ) and Q = (, ). The point R on the x-axis such that PR + RQ is the minimum is Option 5,0 Option,0 Option (, 0) R can be found out by join p to image of a in x-axis p(, ) PR + R Q PR + PQ PR + R Q PR + PQ R minimises the distances eqn. to PQ y+ x- = - 5 Coordinates of R,0
Q.No. 8 The co-ordinates of a point P on the line x - y + 5 = 0 such that PA-PB is maximum where A is (4,-) and B is (,-4) will be : Option (, 7) Option (-,-7) Option (-, 7) Option 4 (0, 5) Correct Answer AP'-BP' AB Equality holds P' P Eqn. to AB x - y = 6 intersects x - y + 5 = 0 at (-, 7) Q.No. 9 A right angle triangle ABC having C a right angle has AC = a and BC = b units. The points A and B slide along the cartesian axes (A on x-axis and B on y-axis). Then the locus of C is : Option by ± ax =0 Option ay ± bx =0 Option x + y = 0 Option 4 xy = ab We have a + b = a + b In BB'C b - h = b'-k and a - k = a'-h ( ) ( ) ( ) ( ) a' + b' = b - h + k + a + k + h a + b = b - h + k + a - k + h by ± ax = 0 [after replacing h = x, k = y]
Q.No. 0 Option ap+ bq+ c acos θ + bsin θ Option aq+ bp+ c acos θ + bsin θ Option aq+ bq+ c asin θ + bcos θ Option 4 aq+ bp+ c asin θ + bcos θ A line is drawn through the point A(p, q) in the direction θ to meet the line, ax + by + c = 0 in B, then AB = Coordinates of B (p + r cos θ, q + r sin θ), B satisfies ax + by + c = 0 (p + r cos θ)a + b(q + r sin θ) + c = 0 ap+ bq+ c r= a cos θ + b sin θ Considering the opposite direction ap+ bq+ c r= acos θ + bsin θ Q.No. The set of values of b for which the origin and the point (, ) lie on the same side of the straight line a x + aby+ = 0 a R, b > 0 are : Option b (,4) Option b (0,) Option b [0,] Option 4 b [, ] Correct Answer (a.0 + ab.0 + ) (a + ab + ) >0 a + ab + >0 For above to hold b - 4<0 b (-, ), b>0 Q.No. Option Option Option The acute angle between two straight lines passing through the point M(-6, -8) and the points in which the line segment x + y + 0 = 0 enclosed between the co-ordinate axes is divided in the ratio : : in the direction from the point of its intersection with the x-axis to the point of intersection with the y-axis is : π π 4 π 6
Option 4 π Correct Answer AP : PQ : QB = : : m (-6, -8) -0-0 P, 5 5 P (-4,-) Q (-,-6) -+8 slopepfpm= = -4+6-6+8 slopepf QM= = -+6 AnglebetweenPMandQM 0 π θ=45 = 4 - + =tanθ Q.No. The equation of the sides of a square whose each side is of length 4 units and centre is (, ). Given that one pair of sides is parallel to x - 4y = 0 Option x - 4y + = 0, x - 4y - 9 = 0, 4x + y + = 0, 4x + y - 7 = 0 Option x - 4y - 5 = 0, x - 4y + 5 = 0, 4x + y + = 0, 4x + y - 7 = 0 Option x - 4y + = 0, x - 4y - 9 = 0, 4x + y + = 0, 4x + y - 8 = 0 Option 4 none Line through (, ) parallel x - 4y is x - 4y + = 0 Let x - 4y + d = 0 be eqn. to AD or BC. d- = d - =0 d = or -9 5 Eqn. to AD or BC are x - 4y - 9 = 0 and x - 4y + = 0
Line through (, ) x - 4y = 0 and 4x + y - 7 = 0 Eqn. to other sides are 4x + y + = 0 or 4x + y - 7 = 0 Q.No. 4 n If,,, β, β, β are the values of n for which - n r x is divisible by - x The triangle having vertices (, β ),(, β )and(, β ) cannot be. Option An isosceles triangle Option A right angled isosceles triangle Option A right angled triangle Option 4 An equilateral triangle Correct Answer 4 n- n- r x is divisible by r x r=0 r=0 n -( ) - n r x x = - x r=0 n - r=0 x r n n -x = - x n r=0 -x -x + x = - - n x x + x For perfect division, n = odd positive intege Since vertices of triangle formed is all rational, can t be equilateral. r=0 r, then Q.No. 5 Consider the straight line ax + by = c, where a, b, c R + this line meets the coordinates axes at A and B respectively. If the area of the OAB, O being origin, does not depend upon a, b and c, then Option a, b, c are in AP Option a, b, c are in GP Option a, b, c are in HP Correct Answer Areaof OAB= OA OB c c = b a c = ab c Areaof OABisindependentof,G.Pinonewaytodoit. ab
Q.No. 6 ABC is an equilateral triangle such that the vertices B and C lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them, then the length of a side of the equilateral triangle is Option 8 Option 88 Option 4 7 Correct Answer Let the length of side be d EB= d -4 CF= d - EB-CF =BD BD = BC - DC ( ) d -4 - d - = d -6 d -4 + d - - d -4 - d - = d -6 d +6-4 - = d -4 - d - (d + 6) = 4(d 4-0d + 64) 4 4 7 d - d =0 d = Q.No. 7 All the points lying on or inside the triangle formed by the points (, ), (5, 6) and (-, ) satisfy Option x+y 0 Option x+ y + 0 Option x+y - 0 Option 4 - x + 0 For the in equation to be satisfied by all points inside or on the triangle, corresponding line won t intersect with the triangle formed. Position of all vertices must be on a single side of the lines. L = x + y L (,),L (5,6),L (-,) 0i.e x+y 0 is satisfiedby allpoints. L (,) =. +. - <0 not satisfied. Other lines are satisfied.
Q.No. 8 Given two straight lines x - y - 7 = 0 and x - y + = 0. Equation of a line which divides the distance between them in the ratio : can be and parallel to them. Option x - y - = 0 Option x - y - = 0 Option y = x Option 4 x - y + = 0 Distance between L and L 0 = = k+k k=, k= Let x - y + d = 0 be L d+7 = d =-or - Eqn. to L : x - y - = 0 Q.No. 9 A light beam emanating from the point A(, 0) reflects from the straight line x + y - 6 = 0 and then passes through the point B(4, ). The equation of the reflected beam is : Option x - y + = 0 Option x + y - = 0 Option x + y - 5 = 0 Option 4 x - y + 5 = 0 Let C (x, 6 - x) slopeofcd= tan ACD = tan DCB ( ) ( ) 6-x-0 6-x- - - x- = x-4 6-x-0 6-x- +. + x- x-4 x+ -x = x = ; y = 4 or x = 5; y = -4 Eqn. of AB : 7x + y - = 0
Point(5, -4) lies on AB; it can t be considered. The diagram will make clear. Required incident point (, 4) Eqn. to reflected ray is x - y + =0 Illustration of (5, -4): Clearly, the incident ray is passing through (4, ), which is not to be taken. Q.No. 40 x y x y x y If + =isalinethroughtheintersectionof + =and + =andthelengthsof c d a b b a The perpendiculars drawn from the origin to these lines are equal to these lines are equal in lengths then : Option + = + a b c d Option - = - a b c d Option + = + a b c d Option 4 none x y x y + =+ λ + - =0is equivalent to a b b a x y + = c d λ λ + + a b + = b a λ = -() c d Equating the value of λ; we get - - a c c b = () - - b d d a Also distance from origin is equal: - - + a c b d + = + = () a b c d - + b d a c - - + c b d a Similarly: = () - + d a c b From(ii)and(iii): - = - and + - + =0 a b d c a b c d I) + = + a b c d
Q.No. 4 Equation of a straight line passing through the point (4, 5) and equally inclined to the lines, x = 4y + 7 and 5y = x + 6 is Option 9x - 7y = Option 9x + 7y = 7 Option 7x + 9y = 7 Option 4 7x - 9y + 7 = 0 The line are inclined to 5y = x + 6-4 L and x = 4y + 7 L so, they would be parallel to angle bisectors of L and L. Equation to angle bisectors: x-5y+ 6 x-4y-7 = ± 5 x + 7y + = 0 and 99x - 77y - 6 = 0 Hence, lines passing through (4, 5) are 7x + 9y = 7 and 9x - 7y = Q.No. 4 Let u ax+ by + a b = 0, v bx- ay+ b a = 0, ab R be two straight lines. The equation of the bisectors of the angle formed by k u - k v = 0 and k u + k v = 0 for non zero real k and k are : Option u = 0 Option k u + k v = 0 Option k u - k v = 0 Option 4 v = 0 u ( ax + by + a b) = 0 v ( bx- ay+ b a)=0 Clearly u v ( ax by a b) ( bx ay b a ) so, X + + ; Y - + In coordinate system XY, the new lines are K X - K Y = 0 and K X + K Y = 0 Eqn. to bisectors: KX-KY KX-K Y = ± X=0 or Y=0 K +K K +K Hence, two lines are u = 0 and v = 0 Q.No. 4 If the equation, x + k xy - y - x - 4y - = 0 represents a pair of lines then the value of k can be: Option Option 5 Option - Option 4-5 Apply condition for a pair of lines. k = -5,
Q.No. 44 Assertion: The lines a x + b y + c = 0, a x + b y + c = 0, a x + b y + c = 0 are a b c concurrent if a b c = 0 a b c Reason: The area of the triangle formed by three concurrent lines must be zero. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion. Option If assertion is true but reason is false. Option 4 If assertion is false but reason is true. Lines are: ai x + bi y + ci = 0, i =,, b c -b c a c -a c a b -b a a b -b a bc -bc ac -ac Areaof = mod =0 a b -b a a b -b a b c -b c a c -a c a b -a b a b -a b a b c a b c a b c =0 Q.No. 45 Assertion: Each point on the line y - x + = 0 is equidistant from the lines 4y + x - = 0, y + 4x - 4 = 0. Reason: The locus of a point which is equidistant from two given lines is the angular bisector of the two lines. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion. Option If assertion is true but reason is false. Option 4 If assertion is false but reason is true. Angle bisectors: 4 x+y-4 = x+4y - x - y - = 0 or 7x + 7y - 6 = 0 Assertion- True Reason- true and correct explanation Q.No. 46 Option Option Option Assertion: if (a x + b y + c ) + (a x + b y + c ) + (a x + b y + c ) = 0, then lines a x + b y + c = 0, a x + b y + c = 0, a x + b y + c = 0 can not be parallel. Reason: If sum of three straight lines equation is identically zero then they are either concurrent or parallel. If both assertion and reason are correct and reason is the correct explanation of assertion. If both assertion and reason are true but reason is not the correct explanation of assertion. If assertion is true but reason is false.
Option 4 If assertion is false but reason is true. Correct Answer 4 (a x + b y + c ) + (a x + b y + c ) + (a x + b y + c ) = 0 i.e. ai x + bi y + c i = 0 are concurrent and can not be parallel. Assertion-true Reason-false Q.No. 47 Assertion: the four straight lines given by x + 7xy - y = 0 and x + 7xy - y - x + 7y - = 0 are the sides of a square. Reason: The lines represented by general equation of second degree ax + hxy + by + gx + fy + c = 0 are perpendicular if a + b = 0 Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion. Option If assertion is true but reason is false. Option 4 If assertion is false but reason is true. Correct Answer x + 7xy - y = 0 and x + 7xy - y - x + 7y - = 0 gives four straight lines x + 4y = 0; 4x - y = 0; x + 4y - = 0 and 4x - y + = 0. They are pair of parallel lines having same distance between them. Assertion-True Reason-True, but not correct explanation. Q.No. 48 Assertion: The diagonals of parallelogram formed by the lines ax + by + c = 0, a x + b y + c = 0, ax + by + c = 0, a x + b y + c = 0 will be perpendicular if aa + bb = 0 Reason: the diagonals of rhombus are always perpendicular. Option If both assertion and reason are correct and reason is the correct explanation of assertion. Option If both assertion and reason are true but reason is not the correct explanation of assertion. Option If assertion is true but reason is false. Option 4 If assertion is false but reason is true. Correct Answer 4 ax + by + c = 0, a x + b y + c = 0, ax + by + c = 0, a x + b y + c = 0 if aa + bb = 0, they are sides of rectangle. Assertion-False Reason-True Passage Text Let A(0,β), B(-, 0) and C (, ) be the vertices of a triangle then Q.No. 49 Angle A of the triangle ABC will be obtuse if β lies in Option (-, ) Option 5, Option -,, Correct Answer
Eqn. to BC: y - x - = 0 A don t lie on BC B A > 90 0 b + c -a < 0 bc. b + c -a < 0 +(- β ) +4+ β -0< 0 bc β (-,), β β -, β Q.No. 50 If I, is the interval of values of β for which A is obtuse and I be the interval of value of β for which A is largest angle of ABC, then Option I = I Option I is a subset of I Option I is a subset of I Correct Answer I I because, even in acute or right triangle, A may be greatest angle. Q.No. 5 All the values of β for which angle A of the triangle ABC is largest lie in interval. Option (-, ) Option -,, Option -,, 6 Correct Answer Case I: A is obture: β -,, (i) Case II: A = 90 0 β =-, (ii) Case III: A is greatest in acute ABC a> b and a> c 0> +(- β ) and 0 > + β β - β -8 < 0 and β < 6 (iii) and β Combining all cares:
-,, 6 β Passage Text Q.No. 5 Option Option Let OX and OY be two fixed lines inclined at a constant angle, A variable line cuts OX at P and OY at Q. From P and Q perpendicular PM and PN are drawn to OY and OX respectively. Let the axes be chosen as OX and OY. If OP = a, OQ = b, then equation of PQ must be x y + = acos bsin x y + = asin bcos Option x y + = a b Correct Answer In XY coordinate, equation to PQ x y + = a b Q.No. 5 Option Option Option The equation of MN must be x y + = acos bsin x y + = bcos acos x y + = asin bsin Correct Answer M a cos N b cos Eqn. to MN:
x y + =L bcos acos Q.No. 54 If AB passes through a fixed point (h, k) then MN passes through a fixed point Option ( kcos, h sin ) Option ( kcos, h cos ) Option ( ksin, h cos ) Correct Answer Let (y- k) = m(x - b) be line AB and (h, k) be the fixed point. mh-k M ( k- mh)cos ;N cos m x my MN= + = ( k- mh)cos ( mh- k)cos ( x-kcos )+ mh ( cos - y)=0passes through ( kcos, h cos ) Q.No. 55 Match the values of k for which origin and (k, ) lie No. Column A Column B Column C Id of Additional Answer (A) In the same angle formed (P)(-4, ) A-R by lines x - y + = 0 and x + y + 5 = 0 (B) Opposite angles formed by lines x - y + = 0 and x + y + 5 = 0 (Q) (-,-4) B-Q (C) Adjacent angles formed by lines x - y + = 0 and x + y + 5 = 0 (R) (, ) (A)(0, 0) and (k, ) lie in the same angle ( k-6+) > 0 and5( k++5) > 0 k > (B)(0, 0) and (k, ) lie in opposite angle ( k-6+) < 0 and5( k++5) < 0 k < -4 (C)(0, 0) and (k, ) lie in adjacent angle ( k-6+)() < 0 and5( k ++5) > 0 Or ( k-6+)> 0 and5( k ++5) < 0 C-P ( k-)( k+4) < 0 x (-4,)
Q.No. 56 A lines cuts X-axis at A and Y-axis at B such that AB = l. Match the following loci: No. Column A Column B Column C Id of Additional Answer (A) Circumcentre of l A-Q triangle ABC (P) x + y = 9 (B) Orthocentre of l B-R triangle ABC (Q) x + y = 4 (C) Incentre of (R) x + y = 0 C-S triangle ABC 4 (D) Centroid of triangle ABC (S) y = x D-P (A) Circumcentre (u, k) (0, b) a + b + l 4h + 4k = l (B) origin is always orthocentre. x + y = 0 l.0+ ab. +0. a l.0+0. b+ ba. (C) Incentre, + + + + l a b l b a x = y (D) centroid (x, y) x = and y = b l x + y = 9 Q.No. 57 For 0 θ < π, if the point (cos θ,sin θ ) lies in the angle between the lines y= x- in which origin lies, then θ lies in the interval of length k π, then k must be 000 Lines are: y = x - and y + x - = 0 (cos θ,sin θ ) lies in the region of (0, 0) (sin θ -cos θ +) > 0and(sin θ +cos θ -) < 0 π - π sin θ - > and sin θ + 4 4 < π π π θ, and θ, π 4 π π θ, π = π = k k Q.No. 58 The co-ordinates of A, B, C are (6, ),(-, 5), (4,-) respectively. For any point P(x, y) if x+ y- the ratio of the areas of the triangles PBCand ABCis. λ Then the numerical quantity λmust be equal to 0007 A (6,);B (-,5); C (4,-);P ( x, y )
6 A r ( ABC)=mod- 5 =49 4 - x y A r ( PBC)=mod- 5 4 - = 7 x+7y -4 A r( PBC) x+ y- x+ y- = = A r( ABC) 7 λ λ = Q.No. 59 Two sides of a rhombus lying in the first quadrant are given by x - 4y = 0 and x - 5y = 0. The length of the longer diagonal is. If the equations of other two sides 80 468 arex-4 y=- andx-5 y= thenthenumericalquantity k shouldbe k k 00 OB = OA : x - 4y = 0 OC = x - 5y = 0-5 4 tan θ = = 6 + 56 0 In =OCB: a - cos(80- θ)= a -56 a - = 65 a 6 0 a = A ( acos, a sin ) tan = 4 4 0 8 0 B, 55 55 0 0 7 0 C, 4 4 4 54 D, 0 0
Eqn. to CB: -98 0-80 x-4 y = = 4 0 Eqn. to AB: 468 x-5 y= =0 0 k Q.No. 60 The base of a triangle passes through a fixed point (, ) and its sides are bisected at right angle by the lines y - 8xy - 9x = 0. If the locus of its vertex is a circle of radius λ.thenthenumericalquantitylmustbeequal. 004 Let A (h, k) BC y - = m(x - ) of y + x = 0 of (y - 9x) = 0 Eqn. to AB: 9y + x = 9k + h Eqn. to AC: y - x = k - h 9 k+ h 8 k+9h F, is midpoint of AB 8 8 9k-40h 40 k+9h B, 4 4 h-k k-h E, is midpoint of AC C (- k,- h ) Form eqn. of BC, eliminating m 9k-40h 40 k+9h - - 4 = 4 -k- -h- (9 k -40 h -4) 40 +9-4 = k h k+ h+ 4 h +4 k +9h- k =0 4 whichisacircleofradius=