Question

if centroid of triangle formed by (p,q), (q,1), (1,p) is the origin , then p^3+q^-3pq

Answer 1

Answer: Value of  p³ + q³ - 3pq  is  -1

Step-by-step explanation:

Coordinate of first vertex of triangle = (p, q)

Coordinate of second vertex of triangle = (q, 1)

Coordinate of third vertex of triangle = (1, p)

Coordinate of centroid of triangle =  $(\frac{x_1 +x_2 + x_3}{3} , \frac{y_1 +y_2 + y_3}{3})$

Coordinate of centroid of triangle =  $(\frac{p+q+1}{3} , \frac{q+p+1}{3})$

ATQ,

Centroid ≡ Origin

(0,0) ≡   $(\frac{p+q+1}{3} , \frac{q+p+1}{3})$

Therefore,

p + q +1 = 0

p + q = -1

( p + q )³  = p³ + q³ + 3pq(p + q)

(-1)³  =  p³ + q³ + 3pq(-1)

-1 = p³ + q³ - 3pq

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Answer 2

The value of p³+q³-3pq is -1.

Given,

The centroid of the triangle formed by (p,q), (q,1), and (1,p) is the origin.

To Find,

The value of p³+q³-3pq.

Solution,

The centroid of the triangle is the origin (0,0).

The vertices of the triangle are (p,q), (q,1), and (1,p).

So,

(p+q+1)/3 = 0

p+q = -1

Now,

(p+q)³ = p³+q³+3pq(p+q)

Substituting the value of p+q = -1

(-1)³ = p³+q³+3pq(-1)

p³+q³-3pq = -1

Hence, the value of p³+q³-3pq is -1.

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