Question

Solution

Correct option is Let O(o, o) be the origin, then

(OA)2 = (OB)2 = (OC)2

⇒     1 + a2 = aa4 = a+ a4

⇒           a2 = 1 ⇒ a = ± 1 but a = 1 is not possible as the points A, B, C will not make a triangle so a = –1, then the vertices of the

triangle are A(1, –1), B(–1, 1), C(1, 1) and the cendroid is SIMILAR QUESTIONS

Q1

ABCD is a rectangle with A(–1,2),B(3,7) and AB : BC = 4:3. If isthe center of the rectangle then the distance of p from each corner is equal to

Q2

If a (2,0) and (0,2) are given points and p is a point such that PA:PB = 2:3 then the locus of p passes through the point (a,a) for

Q3

A(1, 3), B(3, 7) & C(7, 15) are three points. P is the midpoint of ABQ is the midpoint of BC. Locus of a point R which satisfies (PR)2 – (QR)= (AC)2 is

Q4

Given the point A(0, 4) and B(0, –4), the equation of the locus of the pointp(x, y) such that |AP – BP| = 6 is

Q5

Coordinate (x, y) of a point P satisfy the relation 3x + 4= 9, y = mx + 1. The number of integral value of m for which the x-coordinate of p is also an integer is

Q6

The point A(2, 3), B(3, 5), C(7, 7) and D(4, 5) are such that

Q7

Q, R and are the points on the line joining the points P(a, x) and T(b, y) such that PQ = QR = RS = ST.

Q8

The line joining A(bcos α, bsin α) and B(acos β, asin β) is produced to point M(x, y) so that AM : MB = b : a, then Q9

OPQR is square and M, N are the middle points of the sides PQ and QRrespectively then the ratio of the areas of the square and the triangle OMNis

Q10

If px1x2….xi,….and q y1y2,…y… are in A.P. with common difference a and b respectively, then locus of the center of mean position of the point Ai (xi, yi), = 1, 2 …n is