To find the coordinates of point $P$ that satisfies the given conditions, we'll need to use the information that:

1. The x-coordinate of $P$ is twice its y-coordinate.
2. $P$ is equidistant from points $Q(2, -5)$ and $R(-3, 6)$.

Step 1: Express $P$ in Terms of One Variable

Let the coordinates of $P$ be $(x, y)$.

Given that the x-coordinate is twice the y-coordinate, we have:

$x = 2y$

Thus, the coordinates of $P$ can be expressed as $(2y, y)$.

Step 2: Set Up the Distance Equations

The distance from $P$ to $Q(2, -5)$ is:

$PQ = \sqrt{(2y - 2)^2 + (y + 5)^2}$

The distance from $P$ to $R(-3, 6)$ is:

$PR = \sqrt{(2y + 3)^2 + (y - 6)^2}$

Since $P$ is equidistant from $Q$ and $R$, we have:

$PQ = PR$

Step 3: Equate and Simplify the Distances

$\sqrt{(2y - 2)^2 + (y + 5)^2} = \sqrt{(2y + 3)^2 + (y - 6)^2}$

Square both sides to eliminate the square roots:

$(2y - 2)^2 + (y + 5)^2 = (2y + 3)^2 + (y - 6)^2$

Expand each squared term:

• Left Side:

  $(2y - 2)^2 = 4y^2 - 8y + 4$ $(y + 5)^2 = y^2 + 10y + 25$

  Combining these gives:

  $4y^2 - 8y + 4 + y^2 + 10y + 25 = 5y^2 + 2y + 29$

• Right Side:

  $(2y + 3)^2 = 4y^2 + 12y + 9$ $(y - 6)^2 = y^2 - 12y + 36$

  Combining these gives:

  $4y^2 + 12y + 9 + y^2 - 12y + 36 = 5y^2 + 45$

Step 4: Solve for $y$

Equate the two expanded expressions:

$5y^2 + 2y + 29 = 5y^2 + 45$

Subtract $5y^2$ from both sides:

$2y + 29 = 45$

Subtract 29 from both sides:

$2y = 16$

Divide by 2:

$y = 8$

Step 5: Find $x$

Using the relation $x = 2y$, substitute $y = 8$:

$x = 2 \times 8 = 16$

Solution

Thus, the coordinates of $P$ are $(16, 8)$.

Conclusion

• The point $P$ that satisfies the given conditions has coordinates $(16, 8)$.

If you have any questions or need further clarification, feel free to ask!

Related Questions

1. How do you find the midpoint between two points?
2. What is the significance of a point being equidistant from two others in geometry?
3. How can you derive the condition for equidistance from two points?
4. What are the properties of points on the perpendicular bisector of a line segment?
5. How does the equation of a line change if the slope is zero?
6. What are some applications of distance formulas in real-world scenarios?
7. How do you determine the distance from a point to a line?
8. How can you use the concept of loci to solve geometric problems?

Tip: Understanding geometric properties like equidistance can help solve complex problems by breaking them into simpler components.