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


Correct option is


Since (AB)2 + (AC)2 = 50 = (BC)2 the lines AB and AC are at right angle.

It is given that all side are integers and since the triangle is right angled, they from a Pythagorean tripled, so one leg must be even. As two sides are primes, both cannot be the legs of the right angled triangle. 

So, let AC = p, a prime, BC = p + 50 and AB = a (even)

Then     a2 = (p + 50)2 – p2 = 100 (p + 25)


Since a is least possible integer and p is a prime

We get a = 60 when p = 11.

So smallest value of the third side is 60.

           S = 60 + 11 + 61 = 132

           S = 60 × 132 = 7920.




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


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


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


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


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


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


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

                          3x2 + 3y2 + 2xy = 2


                 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


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


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