Answer :
To find the equation of the locus of points that are equidistant from the points \((-1, -1)\) and \( (4, 2) \), follow these steps:
1. Identify the given points:
- Let \( A(-1, -1) \) and \( B(4, 2) \) be the two points.
2. Define the coordinates of the point \(P(x, y)\):
- Let \( P(x, y) \) be any point on the locus. Thus, \( P(x, y) \) is equidistant from both \( A \) and \( B \).
3. Write the expressions for distances:
- The distance from \( P \) to \( A \) is given by:
[tex]\[ PA = \sqrt{(x + 1)^2 + (y + 1)^2} \][/tex]
- The distance from \( P \) to \( B \) is given by:
[tex]\[ PB = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
4. Set the distances equal:
[tex]\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
5. Eliminate the square roots by squaring both sides:
[tex]\[ (x + 1)^2 + (y + 1)^2 = (x - 4)^2 + (y - 2)^2 \][/tex]
6. Expand both sides:
- Left side:
[tex]\[ (x + 1)^2 + (y + 1)^2 = x^2 + 2x + 1 + y^2 + 2y + 1 \][/tex]
[tex]\[ = x^2 + y^2 + 2x + 2y + 2 \][/tex]
- Right side:
[tex]\[ (x - 4)^2 + (y - 2)^2 = x^2 - 8x + 16 + y^2 - 4y + 4 \][/tex]
[tex]\[ = x^2 + y^2 - 8x - 4y + 20 \][/tex]
7. Simplify and equate both sides:
[tex]\[ x^2 + y^2 + 2x + 2y + 2 = x^2 + y^2 - 8x - 4y + 20 \][/tex]
8. Cancel out \( x^2 \) and \( y^2 \) from both sides:
[tex]\[ 2x + 2y + 2 = -8x - 4y + 20 \][/tex]
9. Combine like terms:
[tex]\[ 2x + 2y + 2 + 8x + 4y = 20 \][/tex]
[tex]\[ 10x + 6y + 2 = 20 \][/tex]
10. Isolate the constant term:
[tex]\[ 10x + 6y + 2 - 2 = 20 - 2 \][/tex]
[tex]\[ 10x + 6y = 18 \][/tex]
11. Simplify the equation by dividing through by 2:
[tex]\[ 5x + 3y = 9 \][/tex]
Thus, the equation of the locus of the point equidistant from \( (-1, -1) \) and \( (4, 2) \) is:
[tex]\[ \boxed{y = 3 - \frac{5x}{3}} \][/tex]
1. Identify the given points:
- Let \( A(-1, -1) \) and \( B(4, 2) \) be the two points.
2. Define the coordinates of the point \(P(x, y)\):
- Let \( P(x, y) \) be any point on the locus. Thus, \( P(x, y) \) is equidistant from both \( A \) and \( B \).
3. Write the expressions for distances:
- The distance from \( P \) to \( A \) is given by:
[tex]\[ PA = \sqrt{(x + 1)^2 + (y + 1)^2} \][/tex]
- The distance from \( P \) to \( B \) is given by:
[tex]\[ PB = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
4. Set the distances equal:
[tex]\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
5. Eliminate the square roots by squaring both sides:
[tex]\[ (x + 1)^2 + (y + 1)^2 = (x - 4)^2 + (y - 2)^2 \][/tex]
6. Expand both sides:
- Left side:
[tex]\[ (x + 1)^2 + (y + 1)^2 = x^2 + 2x + 1 + y^2 + 2y + 1 \][/tex]
[tex]\[ = x^2 + y^2 + 2x + 2y + 2 \][/tex]
- Right side:
[tex]\[ (x - 4)^2 + (y - 2)^2 = x^2 - 8x + 16 + y^2 - 4y + 4 \][/tex]
[tex]\[ = x^2 + y^2 - 8x - 4y + 20 \][/tex]
7. Simplify and equate both sides:
[tex]\[ x^2 + y^2 + 2x + 2y + 2 = x^2 + y^2 - 8x - 4y + 20 \][/tex]
8. Cancel out \( x^2 \) and \( y^2 \) from both sides:
[tex]\[ 2x + 2y + 2 = -8x - 4y + 20 \][/tex]
9. Combine like terms:
[tex]\[ 2x + 2y + 2 + 8x + 4y = 20 \][/tex]
[tex]\[ 10x + 6y + 2 = 20 \][/tex]
10. Isolate the constant term:
[tex]\[ 10x + 6y + 2 - 2 = 20 - 2 \][/tex]
[tex]\[ 10x + 6y = 18 \][/tex]
11. Simplify the equation by dividing through by 2:
[tex]\[ 5x + 3y = 9 \][/tex]
Thus, the equation of the locus of the point equidistant from \( (-1, -1) \) and \( (4, 2) \) is:
[tex]\[ \boxed{y = 3 - \frac{5x}{3}} \][/tex]
