Question

Solution

Correct option is

[a + n – 1, b + 2(– 1)]

Now if  x1 = ay1 = b

Then   x2 = a + 2, x3 = + 4, ….. xn = + (– 1)2

y2 = b + 4, y3 = b + 8, ….. yn = b + (n – 1)4

and the coordinates of G are = [a + n – 1, b + 2(n – 1)]

SIMILAR QUESTIONS

Q1

If the line joining the points A(a2, 1) and B(b2, 1) is divides in the ratio b : a at the pint P whose x-coordinate is 7, their

Q2

If two vertices of a triangle are (3, –5) and (–7, 8) and centroid lies at the pint (–1, 1), third vertex of the triangle is at the point (a, b) then

Q3

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

Q4 are two points whose mid-point is at the origin. is a point on the plane whose distance from the origin is

Q5

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

Q6

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

Q7

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

Q8

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

3x2 + 3y2 + 2xy = 2

Q9

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

Q10

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