Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
