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.
When (a, 0) and (0, a) are the coordinates of A and B respectively.
Now equation of MN perpendicular to AB is
So the coordinates of M are
Therefore, area of the triangle AMN is
Also area of the triangle OAB = a2/2.
So that according to the given condition.
For λ = 1/3, M lies outside the segment OB and hence the required value of λ is 3.
The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p andq are the intercepts of the line L on the new axes, then
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 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