Question

If G be the centroid of a triangle ABC and P be any other point in the plane , prove that PA 2 +PB 2 + PC 2 = GA 2 + GB 2 +GC 2 + 3GP 2

Answer 1

Let the coordinates of A, B and C be (x1.y1), (x2.y2) and (x3.y3)

Then coordinates of centroid G will be (x1­ + x2 + x3)/3, (y1 + y2 + y3)/3 )

Let us the assume that centroid lies on the origin
x1­ + x2 + x3 = 0
y1 + y2 + y3 = 0
         
  Let the coordinates of point O be (x, y)
to be proved:
 OA² + OB² + OC² = GA² + GB² + GC² + 3GO²
          
Now, L.H.S. = OA² + OB² + OC²

(x – x1)² + (y – y1)² + (x – x2)² + (y – y2)² + (x – x3)² + (y – y3)²
= 3x² + 3y² + x1² + x2² + x3² + y1² + y2² + y3² – 2x (x1 + x2 + x3) – 2y(y1 + y2 + y3)
= 3x² + 3y² + x1² + x2² + x3² + y1² + y2² + y3²
          
R.H.S. = GA² + GB² + GC² + 3GO²
          
= (x1 – 0)² + (y1 – 0)² + (x2 – 0)² + (y2 – 0)² + (x3 – 0)² + (y3 – 0)² + 3[(x – 0)²+ (y – 0)²]
= x1² + y1² + x2² + y2² + x3² + y3² + 3(x² + y²)
= 3x² + 3y² + x1² + x2² + x3² + y1² + y2² + y3²
= L.H.S.

Answer 2

Step-by-step explanation:

Let the coordinates of A, B and C be (x1.y1), (x2.y2) and (x3.y3)

Then coordinates of centroid G will be ((x1­ + x2 + x3)/3, (y1 + y2 + y3)/3 )

Let us the assume that centroid lies on the origin

⇒ x1­ + x2 + x3 = 0 and y1 + y2 + y3 = 0

Let the coordinates of point O be (x, y)

We have to prove OA2 + OB2 + OC2 = GA2 + GB2 + GC2 + 3GO2

Now, L.H.S. = OA2 + OB2 + OC2

(x – x1)2 + (y – y1)2 + (x – x2)2 + (y – y2)2 + (x – x3)2 + (y – y3)2

= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32 – 2x (x1 + x2 + x3) – 2y(y1 + y2 + y3)

= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32

R.H.S. = GA2 + GB2 + GC2 + 3GO2

= (x1 – 0)2 + (y1 – 0)2 + (x2 – 0)2 + (y2 – 0)2 + (x3 – 0)2 + (y3 – 0)2 + 3[(x – 0)2 + (y – 0)2]

= x12 + y12 + x22 + y22 + x32 + y32 + 3(x2 + y2)

= 3x2 + 3y2 + x12 + x22 + x32 + y12 + y22 + y32

= L.H.S.