## Question

Two vertices of a triangle are (–1, 4) and (5, 2). If its centroid is (0, –3), find the third vertex.

### Solution

**(–4, –15)**

Let the third vertex be (*x*, *y*) then the co – ordinates of the centroid of triangle are

or *x *= – 4 and *y *= – 15

Hence the third vertex is (–4, –15).

#### SIMILAR QUESTIONS

*OPQR* is square and *M, N* are the middle points of the sides *PQ* and *QR*respectively then the ratio of the areas of the square and the triangle *OMN*is

If *p*, *x*_{1}, *x*_{2}*….x _{i}*,….and

*q y*

_{1},

*y*

_{2},…

*y*… are in A.P. with common difference

_{i }*a*and

*b*respectively, then locus of the center of mean position of the point

*A*(

_{i}*x*),

_{i, }y_{i}*i*= 1, 2 …

*n*is

If α, β, γ are the real roots of the equation *x*^{3} – 3*px*^{3} + 3*qx *– 1 = 0, then the centroid of the triangle with vertices

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If *G* is the centroid and *I* the incentre of the triangle with vertices *A*(–36, 7), *B*(20, 7) and *C*(0, –8), then *GI* is equal to

Consider the point then

A variable straight line of slope 4 intersects the hyperbola *xy* = 1 at two points. The locus of the point which divides the line segment between these two points in the ratio 1 : 2 is

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Find the co – ordinates of a point which divides externally the line joining (1, **–**3) and (**–**3, 9) in the ratio 1 : 3.

Find the area of the pentagon whose vertices are A(1, 1), B(7, 21), C(7, –3), D(12, 2) and (0, –3).