Write an equation whose graph consists of the set of points 
P(x, y, z)
 that are twice as far from 
A(0, −1, 1)
 as from 
B(1, 5, 0).

To find the equation of the set of points \( P(x, y, z) \) that are twice as far from point \( A(0, -1, 1) \) as from point \( B(1, 5, 0) \), we proceed as follows:

Let the distance between the point \( P(x, y, z) \) and \( A(0, -1, 1) \) be denoted by \( d_A \), and the distance between \( P(x, y, z) \) and \( B(1, 5, 0) \) be denoted by \( d_B \).

The distance between two points \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \) in 3-dimensional space is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]

1. The distance from \( P(x, y, z) \) to \( A(0, -1, 1) \) is:

\[
d_A = \sqrt{(x - 0)^2 + (y + 1)^2 + (z - 1)^2} = \sqrt{x^2 + (y + 1)^2 + (z - 1)^2}
\]

2. The distance from \( P(x, y, z) \) to \( B(1, 5, 0) \) is:

\[
d_B = \sqrt{(x - 1)^2 + (y - 5)^2 + (z - 0)^2} = \sqrt{(x - 1)^2 + (y - 5)^2 + z^2}
\]

According to the problem, the point \( P(x, y, z) \) is twice as far from \( A \) as it is from \( B \), so we have the relation:

\[
d_A = 2 d_B
\]

Substituting the expressions for \( d_A \) and \( d_B \) into this equation:

\[
\sqrt{x^2 + (y + 1)^2 + (z - 1)^2} = 2 \sqrt{(x - 1)^2 + (y - 5)^2 + z^2}
\]

Now, square both sides to eliminate the square roots:

\[
x^2 + (y + 1)^2 + (z - 1)^2 = 4 \left( (x - 1)^2 + (y - 5)^2 + z^2 \right)
\]

Expand both sides:

Left-hand side:
\[
x^2 + (y + 1)^2 + (z - 1)^2 = x^2 + y^2 + 2y + 1 + z^2 - 2z + 1 = x^2 + y^2 + z^2 + 2y - 2z + 2
\]

Right-hand side:
\[
4 \left( (x - 1)^2 + (y - 5)^2 + z^2 \right) = 4 \left( x^2 - 2x + 1 + y^2 - 10y + 25 + z^2 \right)
\]
\[
= 4 \left( x^2 + y^2 + z^2 - 2x - 10y + 26 \right) = 4x^2 + 4y^2 + 4z^2 - 8x - 40y + 104
\]

Now, equate the two expanded expressions:

\[
x^2 + y^2 + z^2 + 2y - 2z + 2 = 4x^2 + 4y^2 + 4z^2 - 8x - 40y + 104
\]

Move all terms to one side of the equation:

\[
x^2 + y^2 + z^2 + 2y - 2z + 2 - 4x^2 - 4y^2 - 4z^2 + 8x + 40y - 104 = 0
\]

Simplify the expression:

\[
-3x^2 - 3y^2 - 3z^2 + 8x + 42y - 2z - 102 = 0
\]

Finally, divide through by \(-3\) to simplify further:

\[
x^2 + y^2 + z^2 - \frac{8}{3}x - 14y + \frac{2}{3}z + 34 = 0
\]

This is the equation of the set of points that are twice as far from \( A(0, -1, 1) \) as from \( B(1, 5, 0) \).

---

Do you want details or have any questions about this result?  
Here are some related questions to expand your understanding:

1. How would the equation change if the points \( A \) and \( B \) were in different locations?
2. What is the geometric shape represented by the given equation?
3. Can this problem be extended to higher-dimensional spaces?
4. How do you find the center and radius of a sphere from a general quadratic equation?
5. How would you solve the equation numerically for specific points?

**Tip:** Always check whether squaring both sides of an equation introduces extraneous solutions when dealing with distances.

Write an equation whose graph consists of the set of points P(x, y, z) that are twice as far from A(0, −1, 1) as from B(1, 5, 0).

Explore the equation for points P(x, y, z) that are twice as far from A(0, -1, 1) as from B(1, 5, 0). Learn how to use the 3D distance formula and algebraic techniques to solve this problem step-by-step. Ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation