Question

If O is the origin and the coordinates of A and B are (51, 65) and (75, 81) respectively. then  is equal to

Solution

Correct option is

9090

    (OA)2 = (51)2 + (65)2, (OB)2 = (75)2 + (81)2

    (AB)2 = (75 – 51)2 + (81 – 65)2

And from triangle OAB

   

So 

SIMILAR QUESTIONS

Q1

α is root of the equation x2 – 5x + 6 = 0 and β is a root of the equation x2– x – 30 = 0, then coordinates  of the point P farthest from the origin are

Q2

 are two points whose mid-point is at the origin.  is a point on the plane whose distance from the origin is

Q3

Locus of the point P(2t2 + 2, 4t + 3), where t is a variable is

Q4

If the coordinates of An are (n, n2) and the ordinate of the center of mean position of the points A1A2, … An is 46, then n is equal to

Q5

Area of the triangle with vertices A(3, 7), B(–5, 2) and C(2, 5) is denoted by Δ. If ΔA, ΔBΔC denote the areas of the triangle with vertices OBC, AOC and ABO respectively, O being the origin, then

Q6

If the axes are turned through 450. Find the transformed from the equation

                          3x2 + 3y2 + 2xy = 2

Q7

                 A1A2A3, …. An are points in a plane whose coordinates are (x1y1), (x2y2), (x3y3) …, (xnyn) respectively

A1Ais dissected at the point G1GA3 is divided in the ratio  1 : 2 at G2GA4 is divided in the ratio 1 : 4 at G4, and so on until all n points are exhausted. The coordinates of the final point G so obtained are

Q8

If x1 = ay1 = bx1x­2 …. xn and y1y2 …. yn from an ascending arithmetic progressing with common difference 2 abd 4 respectively, then the coordinates of G are

Q9

Let the sides of a triangle ABC are all integers with A as the origin. If (2, –1) and (3, 6) are points on the line AB and AC respectively (lines AB andAC may be extended to contain these points), and length of any two sides are primes that differ by 50. If a is least possible lengths of the third side and S is the least possible perimeter of the triangle then aS is equal to

Q10

Vertices of a triangle are (0, 0), (41a, 37) and (–37, 41b) where a and bare the roots of the equation. 3x2 – 16x + 15 = 0. The area of the triangle is equal to