Question

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

Solution

Correct option is

648

We have 

⇒ a + a2 – 20 = 0 ⇒ a = –5 or 4

⇒ a2 + a3 = 80 which holds for a = 4.

So, the given points are A(5, 3), B(17, 15), C(65, 63) and

Coordinates of F are  and

           (BF)2 = (35 – 17)2 + (33 – 15)2

                     = (18)2 + (18)2 = 648

SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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

                          3x2 + 3y2 + 2xy = 2

Q4

                 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

Q5

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

Q6

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

Q7

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

Q8

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

Q9

If O is the origin and An is the point with coordinates (n, n + 1) then (OA1)2 + (OA2)2 + …. + (OA7)2 is equal to

Q10

Given two points A(–2, 0), and B(0,4),  is a point with coordinates (x, x), x ≥ 0P divides the joint A and B in the ratio 2 : 1. C and D are the mid-point of BM and MA respectively

1:- Area of the ΔAMB is minimum, if the coordinates of M are