Question

Let $A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$. There exists a point $X$ and a constant $k$ such that for any point $P$,

\[PA^2 + PB^2 + PC^2 = 3PX^2 + k.\]Find the constant $k$.

I've tried, but got that k = 4x + 18 2/3, which is definitely wrong. I thought to take out the "4x", but its still wrong.

Why is no one helping :(

Answer 1

k=32

Explanation:

Given the points:

`A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.`

The first step is to find the Centroid of the triangle.

Centroid, X

`=\left((x_1+x_2+x_3)/(3) ,(y_1+y_2+y_3)/(3) \right)\\=\left((4+6+(-1))/(3) ,(-1+2+2)/(3) \right)\\=\left((9)/(3) ,(3)/(3) \right)=(3,1)`

Next, let P be a point (x,y)

Using the distance formula,
`√((x_2-x_1)^2+(y_2-y_1)^2)`

`PA^2=(x-4)^2+(y-(-1))^2\\PB^2=(x-6)^2+(y-2)^2\\PC^2=(x-(-1))^2+(y-2)^2\\PX^2=(x-3)^2+(y-1)^2\\`

On Substitution into:
`PA^2 + PB^2 + PC^2 = 3PX^2 + k`

`(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k`

Let us simplify the LHS first

`\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62`

Also, the Right Hand Side

`RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k`

Therefore:

`3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32`