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
Let B, C, D, E 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 (x, y) 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 (x, y) 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 B, P is the mid-point ofAB (Fig). P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3is 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 OAB, O being the origin, with right angle at N. M 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 A (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