Find the equation of set of points $\mathrm{P}$ such that $\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 k^{2}$, where $\mathrm{A}$ and $\mathrm{B}$ are the points $(3{,}4{,}5)$ and $(-1{,}3,-7)$, respectively.

Step 1: Define coordinates of points

We start by defining the coordinates of the unknown point $P$ as $(x, y, z)$ and the given points $A$ and $B$ with their respective coordinates.

Step 2: Calculate squared distances PA² and PB²

Using the distance formula, the square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$. We apply this formula to find $\mathrm{PA}^2$ and $\mathrm{PB}^2$.

Step 3: Substitute into the given equation

Now, we substitute the expressions for $\mathrm{PA}^2$ and $\mathrm{PB}^2$ into the given equation $\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 k^{2}$.

Step 4: Expand and simplify the equation

We expand all the squared terms and then combine like terms ($x^2$, $x$, $y^2$, $y$, $z^2$, $z$, and constant terms) to simplify the equation.

Step 5: Divide by 2 and rearrange

To further simplify, we divide the entire equation by 2. Then, we move the $k^2$ term to the left side to express the equation in a standard form.

More from this chapter

• The centroid of a triangle ABC is at the point (1,1,1). If the coordinates of A and B are (3,-5,7) and (-1,7,-6), respectively, find the coordinates of the poin
• Show that the points A (1, 2, 3), B (-1, -2, -1), C (2, 3, 2) and D(4,7,6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
• Are the points A (3,6,9), B (10,20,30) and C(25,-41,5), the vertices of a right angled triangle?
• Find the equation of the set of the points P such that its distances from the points A(3,4,-5) and B(-2,1,4) are equal.
• Show that the points P(-2,3,5), Q(1,2,3) and R(7,0,-1) are collinear.

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