## Question

### Solution

Correct option is

1/2n

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. (say)     #### SIMILAR QUESTIONS

Q1

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

Q2

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 Q3

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

Q4

The area enclosed by 2|x| + 3|y≤ 6 is

Q5

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

Q6

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

Q7

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

Q8

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

Q9

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

Q10

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.