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To solve the given problem, we need to find the relationship between the squared distances from an unknown point [tex]\( P \)[/tex] to two fixed points, [tex]\( A(0,1) \)[/tex] and [tex]\( B(-4,3) \)[/tex], given that their difference is 16. Specifically, [tex]\( PA^2 - PB^2 = 16 \)[/tex].
### Step-by-Step Solution:
1. Define the coordinates of the points:
- Point [tex]\( A \)[/tex] has coordinates [tex]\( (0,1) \)[/tex].
- Point [tex]\( B \)[/tex] has coordinates [tex]\( (-4,3) \)[/tex].
2. Let the coordinates of point [tex]\( P \)[/tex] be [tex]\( (x, y) \)[/tex].
3. Express the squared distance from point [tex]\( P \)[/tex] to point [tex]\( A \)[/tex]:
[tex]\[ PA^2 = (x - 0)^2 + (y - 1)^2 = x^2 + (y - 1)^2 \][/tex]
4. Express the squared distance from point [tex]\( P \)[/tex] to point [tex]\( B \)[/tex]:
[tex]\[ PB^2 = (x + 4)^2 + (y - 3)^2 \][/tex]
5. Set up the equation for the difference of squares:
[tex]\[ PA^2 - PB^2 = 16 \][/tex]
6. Substitute the expressions for [tex]\( PA^2 \)[/tex] and [tex]\( PB^2 \)[/tex] into the equation:
[tex]\[ x^2 + (y - 1)^2 - \left( (x + 4)^2 + (y - 3)^2 \right) = 16 \][/tex]
7. Expand the squares:
[tex]\[ x^2 + (y^2 - 2y + 1) - \left( (x^2 + 8x + 16) + (y^2 - 6y + 9) \right) = 16 \][/tex]
8. Combine like terms:
[tex]\[ x^2 + y^2 - 2y + 1 - (x^2 + 8x + 16 + y^2 - 6y + 9) = 16 \][/tex]
9. Simplify the equation by canceling common terms:
[tex]\[ x^2 + y^2 - 2y + 1 - x^2 - 8x - 16 - y^2 + 6y - 9 = 16 \][/tex]
10. Combine the terms on the left:
[tex]\[ -8x + 4y - 24 = 16 \][/tex]
11. Isolate the terms involving [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ -8x + 4y = 40 \][/tex]
12. Divide the entire equation by 4 to simplify:
[tex]\[ -2x + y = 10 \][/tex]
13. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = 2x + 10 \][/tex]
Therefore, the given constraint [tex]\(PA^2 - PB^2 = 16\)[/tex] implies that any point [tex]\( P (x, y) \)[/tex] located on the line defined by the equation [tex]\( y = 2x + 10 \)[/tex] will satisfy the condition.
### Step-by-Step Solution:
1. Define the coordinates of the points:
- Point [tex]\( A \)[/tex] has coordinates [tex]\( (0,1) \)[/tex].
- Point [tex]\( B \)[/tex] has coordinates [tex]\( (-4,3) \)[/tex].
2. Let the coordinates of point [tex]\( P \)[/tex] be [tex]\( (x, y) \)[/tex].
3. Express the squared distance from point [tex]\( P \)[/tex] to point [tex]\( A \)[/tex]:
[tex]\[ PA^2 = (x - 0)^2 + (y - 1)^2 = x^2 + (y - 1)^2 \][/tex]
4. Express the squared distance from point [tex]\( P \)[/tex] to point [tex]\( B \)[/tex]:
[tex]\[ PB^2 = (x + 4)^2 + (y - 3)^2 \][/tex]
5. Set up the equation for the difference of squares:
[tex]\[ PA^2 - PB^2 = 16 \][/tex]
6. Substitute the expressions for [tex]\( PA^2 \)[/tex] and [tex]\( PB^2 \)[/tex] into the equation:
[tex]\[ x^2 + (y - 1)^2 - \left( (x + 4)^2 + (y - 3)^2 \right) = 16 \][/tex]
7. Expand the squares:
[tex]\[ x^2 + (y^2 - 2y + 1) - \left( (x^2 + 8x + 16) + (y^2 - 6y + 9) \right) = 16 \][/tex]
8. Combine like terms:
[tex]\[ x^2 + y^2 - 2y + 1 - (x^2 + 8x + 16 + y^2 - 6y + 9) = 16 \][/tex]
9. Simplify the equation by canceling common terms:
[tex]\[ x^2 + y^2 - 2y + 1 - x^2 - 8x - 16 - y^2 + 6y - 9 = 16 \][/tex]
10. Combine the terms on the left:
[tex]\[ -8x + 4y - 24 = 16 \][/tex]
11. Isolate the terms involving [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ -8x + 4y = 40 \][/tex]
12. Divide the entire equation by 4 to simplify:
[tex]\[ -2x + y = 10 \][/tex]
13. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = 2x + 10 \][/tex]
Therefore, the given constraint [tex]\(PA^2 - PB^2 = 16\)[/tex] implies that any point [tex]\( P (x, y) \)[/tex] located on the line defined by the equation [tex]\( y = 2x + 10 \)[/tex] will satisfy the condition.
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