Question

If a (-5,-1) ,b(-1,0) ,c(1,2) ,d(1,3) and p is any other point such that d = pa^2 +pb^2+pc^2 +pd^2 then the least possible value of d is

Answer 1

Least possible value of 'd' is 34 at (- 1, 1)

Step-by-step explanation:

The given points are

• A (- 5, - 1)

• B (- 1, 0)

• C (1, 2)

• D (1, 3)

Let the coordinates of the point P are (x, y)

Now, d = PA² + PB² + PC² + PD²

= [√{(x + 5)² + (y + 1)²}]² + [√{(x + 1)² + (y - 0)²}]² + [√{(x - 1)² + (y - 2)²}]² + [√{(x - 1)² + (y - 3)²}]²

= (x² + y² + 10x + 2y + 26) + (x² + y² + 2x + 1) + (x² + y² - 2x - 4y + 5) + (x² + y² - 2x - 6y + 10)

= (4x² + 4y² + 8x - 8y) + 42

We put x = - 1 and y = 1

Then d = 4 (- 1)² + 4 (1)² + 8 (- 1) - 8 (1) + 42

= 4 + 4 - 8 - 8 + 42

= 34

∴ the minimum value of d is 34 at (- 1, 1)