Answer :
To solve this problem, we will approach it in two parts, as there are two distinct loci to find.
### Part 1: Locus Equidistant from Points \((-1, -1)\) and \((4, 2)\)
1. Identify the general problem:
- We need to find the set of points \((x, y)\) that are equidistant from both \((-1, -1)\) and \((4, 2)\).
2. Use the distance formula:
- The distance from any point \((x, y)\) to \((-1, -1)\) is \(\sqrt{(x - (-1))^2 + (y - (-1))^2} = \sqrt{(x + 1)^2 + (y + 1)^2}\).
- The distance from any point \((x, y)\) to \((4, 2)\) is \(\sqrt{(x - 4)^2 + (y - 2)^2}\).
3. Set up the equation:
- These two distances must be equal, resulting in the equation:
[tex]\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
4. Simplify the equation:
- To eliminate the square roots, square both sides:
[tex]\[ (x + 1)^2 + (y + 1)^2 = (x - 4)^2 + (y - 2)^2 \][/tex]
- Expand both sides of the equation:
[tex]\[ (x^2 + 2x + 1) + (y^2 + 2y + 1) = (x^2 - 8x + 16) + (y^2 - 4y + 4) \][/tex]
- Combine like terms:
[tex]\[ x^2 + 2x + y^2 + 2y + 2 = x^2 - 8x + y^2 - 4y + 20 \][/tex]
- Cancel out similar terms (\(x^2\) and \(y^2\)):
[tex]\[ 2x + 2y + 2 = -8x - 4y + 20 \][/tex]
- Combine all x and y terms on one side and constant terms on the other:
[tex]\[ 2x + 2y + 2 + 8x + 4y = 20 \][/tex]
[tex]\[ 10x + 6y + 2 = 20 \][/tex]
- Simplify to:
[tex]\[ 10x + 6y = 18 \][/tex]
- Divide through by 2 to make it simpler:
[tex]\[ 5x + 3y = 9 \][/tex]
So, the equation of the locus equidistant from \((-1, -1)\) and \((4, 2)\) is:
[tex]\[ 5x + 3y = 9 \][/tex]
### Part 2: Locus Equidistant from Point \((2, 4)\) and the Y-axis
1. Identify the general problem:
- We need to find the set of points \((x, y)\) that are equidistant from \((2, 4)\) and the Y-axis.
2. Use the distance formula:
- The distance from any point \((x, y)\) to \((2, 4)\) is \(\sqrt{(x - 2)^2 + (y - 4)^2}\).
- The distance from any point \((x, y)\) to the Y-axis is \(|x|\).
3. Set up the equation:
- These two distances must be equal, resulting in the equation:
[tex]\[ \sqrt{(x - 2)^2 + (y - 4)^2} = |x| \][/tex]
4. Simplify the equation:
- Square both sides to remove the square root:
[tex]\[ (x - 2)^2 + (y - 4)^2 = x^2 \][/tex]
- Expand and simplify:
[tex]\[ (x^2 - 4x + 4) + (y^2 - 8y + 16) = x^2 \][/tex]
- Combine like terms and simplify:
[tex]\[ x^2 - 4x + 4 + y^2 - 8y + 16 = x^2 \][/tex]
- Cancel out \(x^2\) terms from both sides:
[tex]\[ - 4x + 4 + y^2 - 8y + 16 = 0 \][/tex]
- Further simplify and collect similar terms:
[tex]\[ y^2 - 8y - 4x + 20 = 0 \][/tex]
So, the equation of the locus equidistant from \((2, 4)\) and the Y-axis is:
[tex]\[ y^2 - 8y - 4x + 20 = 0 \][/tex]
### Part 1: Locus Equidistant from Points \((-1, -1)\) and \((4, 2)\)
1. Identify the general problem:
- We need to find the set of points \((x, y)\) that are equidistant from both \((-1, -1)\) and \((4, 2)\).
2. Use the distance formula:
- The distance from any point \((x, y)\) to \((-1, -1)\) is \(\sqrt{(x - (-1))^2 + (y - (-1))^2} = \sqrt{(x + 1)^2 + (y + 1)^2}\).
- The distance from any point \((x, y)\) to \((4, 2)\) is \(\sqrt{(x - 4)^2 + (y - 2)^2}\).
3. Set up the equation:
- These two distances must be equal, resulting in the equation:
[tex]\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{(x - 4)^2 + (y - 2)^2} \][/tex]
4. Simplify the equation:
- To eliminate the square roots, square both sides:
[tex]\[ (x + 1)^2 + (y + 1)^2 = (x - 4)^2 + (y - 2)^2 \][/tex]
- Expand both sides of the equation:
[tex]\[ (x^2 + 2x + 1) + (y^2 + 2y + 1) = (x^2 - 8x + 16) + (y^2 - 4y + 4) \][/tex]
- Combine like terms:
[tex]\[ x^2 + 2x + y^2 + 2y + 2 = x^2 - 8x + y^2 - 4y + 20 \][/tex]
- Cancel out similar terms (\(x^2\) and \(y^2\)):
[tex]\[ 2x + 2y + 2 = -8x - 4y + 20 \][/tex]
- Combine all x and y terms on one side and constant terms on the other:
[tex]\[ 2x + 2y + 2 + 8x + 4y = 20 \][/tex]
[tex]\[ 10x + 6y + 2 = 20 \][/tex]
- Simplify to:
[tex]\[ 10x + 6y = 18 \][/tex]
- Divide through by 2 to make it simpler:
[tex]\[ 5x + 3y = 9 \][/tex]
So, the equation of the locus equidistant from \((-1, -1)\) and \((4, 2)\) is:
[tex]\[ 5x + 3y = 9 \][/tex]
### Part 2: Locus Equidistant from Point \((2, 4)\) and the Y-axis
1. Identify the general problem:
- We need to find the set of points \((x, y)\) that are equidistant from \((2, 4)\) and the Y-axis.
2. Use the distance formula:
- The distance from any point \((x, y)\) to \((2, 4)\) is \(\sqrt{(x - 2)^2 + (y - 4)^2}\).
- The distance from any point \((x, y)\) to the Y-axis is \(|x|\).
3. Set up the equation:
- These two distances must be equal, resulting in the equation:
[tex]\[ \sqrt{(x - 2)^2 + (y - 4)^2} = |x| \][/tex]
4. Simplify the equation:
- Square both sides to remove the square root:
[tex]\[ (x - 2)^2 + (y - 4)^2 = x^2 \][/tex]
- Expand and simplify:
[tex]\[ (x^2 - 4x + 4) + (y^2 - 8y + 16) = x^2 \][/tex]
- Combine like terms and simplify:
[tex]\[ x^2 - 4x + 4 + y^2 - 8y + 16 = x^2 \][/tex]
- Cancel out \(x^2\) terms from both sides:
[tex]\[ - 4x + 4 + y^2 - 8y + 16 = 0 \][/tex]
- Further simplify and collect similar terms:
[tex]\[ y^2 - 8y - 4x + 20 = 0 \][/tex]
So, the equation of the locus equidistant from \((2, 4)\) and the Y-axis is:
[tex]\[ y^2 - 8y - 4x + 20 = 0 \][/tex]
