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 =  


Correct option is


y = 1 meets x-axis at A (1, 0) and y-axis at B(0, 1) (Fig). 



The coordinates of P are (1/2, 1/2) and PP1 is perpendicular to OA.

Equation of line OP is y = x.









If two vertices of a triangle are (–2, 3) and (5, –1), orthocentre lies at the origin and centroid on the line x + y = 7, then the third vertex lies at


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 (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 = 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.