The point (4, 1) undergoes the following transformation successively.

(i) Reflection about the line y = x

(ii) Translation through a distance 2 units along the positive direction of x-axis.

(iii) Rotation through an anlge π/4 about the origin in the anticlockwise direction  

(iv) Reflection about x = 0

The final position of the given point is



Correct option is

Let BCDE be the positions of the given point A(4, 1) after the transformations (i), (ii), (iii), and (iv) successively (Fig).                                           




The coordinates of B are (1, 4) and that of C are (1 + 2, 4 + 0) i.e. (3, 4). Now if OC makes an angle θ with x-axis, OD make and an angle θ + π/4 with x-axis. If (h, k) denote the coordinates of D. then 

          h = OD cos (θ + 45o), k = sin (θ + 45o)

and OD = OC = 5, sin θ = 4/5, cos θ = 3/5 



Coordinates of D are  and its reflection about

x = 0 in      










If P is a point (xy) on the line, y = –3x such that P and the point (3, 4) are on the opposite sides of the line 3x – 4y = 8, then   



The area enclosed by 2|x| + 3|y≤ 6 is


Let O be the origin, A (1, 0) and B (0, 1) and P (xy) are points such thatxy > 0 and x + y < 1, then


If a line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through and angle 15o, then equation of the line is the new position is


An equation of a line through the point (1, 2) whose distance from the point (3, 1) has the greatest value is


Let 0 < α < π/2 be a fixed angle. If P = (cos θ, sin θ) and Q = (cos (α – θ), sin (α – θ) then Q is obtained from P by


On the portion of the straight line x + y = 2 which is intercepted between the axes, a square is constructed, away from the origin, with this portion as one of its side. If p denotes the perpendicular distance of a side of this square from the origin, then the maximum value of p is 


The line x + y = 1 meets x-axis at A and y-axis at BP is the mid-point ofAB (Fig). P is the foot of the perpendicular from P to OAM1 is that from P1 to OPP2 is that from M1 to OAM2 is that from P2 to OPP3is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn – 1, then OPn =  


The line x + y = a, meets the axis of x and y at A and B respectively. A triangle AMN is inscribed in the triangle OABO being the origin, with right angle at NM and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to.


A line cuts the x-axis at (7, 0) and the y-axis at B(0, –5). A variable linePQ is draw perpendicular to AB cutting the x-axis at P and the y-axis at in θ. If AQ and BP intersect at R, the locus of R is