## Question

### Solution

Correct option is

343

(OA1)2 + (OA2)2 + …. + (OA7)2

= 12 + 22 + 22 + 32 + ….. + 72 + 82

= 2(12 + 22 + …. + 72) – 1 + 82  #### SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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

Q4

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

3x2 + 3y2 + 2xy = 2

Q5

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

Q6

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

Q7

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

Q8

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

Q9

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

Q10

A(a + 1, a – 1), B(a2 + 1, a2 – 1) and C(a3 + 1, a3 – 1) are given points D(11, 9) is the mid-point of AB and E(41, 39) is the mid-point of BC. If F is the mid-point of AC the (BF)2 is equal to