Find the equation of the set of the points $\mathrm{P}$ such that its distances from the points $\mathrm{A}(3{,}4,-5)$ and $\mathrm{B}(-2{,}1{,}4)$ are equal.

Step 1: Define point P and the distance formula

Let the coordinates of point P be $(x,y,z)$. The problem states that the distance from P to A is equal to the distance from P to B. We will use the standard 3D distance formula to express these distances.

Step 2: Set up the equality of distances

We set the distance PA equal to the distance PB. Substitute the coordinates of A $(3{,}4,-5)$ and B $(-2{,}1{,}4)$ into the distance formula. Remember that $z - (-5)$ becomes $z+5$ and $x - (-2)$ becomes $x+2$.

Step 3: Square both sides and expand

To eliminate the square roots, we square both sides of the equation. Then, we expand each squared term using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$.

Step 4: Simplify the equation

Notice that $x^2$, $y^2$, and $z^2$ terms cancel out from both sides of the equation. Combine the constant terms on each side. Then, move all terms to one side to simplify the equation.

Step 5: Rearrange to standard form

Gather all $x$, $y$, and $z$ terms, and constant terms on one side of the equation. This gives us the final linear equation representing the set of points P.

More from this chapter

• Find the equation of set of points P such that PA²+PB²=2 k², where A and B are the points (3,4,5) and (-1,3,-7), respectively.
• 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?
• Show that the points P(-2,3,5), Q(1,2,3) and R(7,0,-1) are collinear.

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