Answer :
Sure, let's solve this problem step by step.
We are given two points \((1, 2)\) and \((2, -3)\) and we need to find the set of all points \((x, y)\) in the plane which are at an equal distance from these two given points.
To start, let's use the distance formula. The distance between any two points \((x_1, y_1)\) and \((x_2, y_2)\) in the plane is given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let's apply this to our problem.
1. The distance from point \((x, y)\) to the point \((1, 2)\):
[tex]\[ d_1 = \sqrt{(x - 1)^2 + (y - 2)^2} \][/tex]
2. The distance from point \((x, y)\) to the point \((2, -3)\):
[tex]\[ d_2 = \sqrt{(x - 2)^2 + (y + 3)^2} \][/tex]
Since we are looking for the points that are equidistant to \((1, 2)\) and \((2, -3)\):
[tex]\[ \sqrt{(x - 1)^2 + (y - 2)^2} = \sqrt{(x - 2)^2 + (y + 3)^2} \][/tex]
To isolate \(y\), we first eliminate the square roots by squaring both sides:
[tex]\[ ((x - 1)^2 + (y - 2)^2) = ((x - 2)^2 + (y + 3)^2) \][/tex]
Now we expand both sides:
[tex]\[ (x - 1)^2 + (y - 2)^2 = (x - 2)^2 + (y + 3)^2 \][/tex]
Expanding each term:
[tex]\[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = (x^2 - 4x + 4) + (y^2 + 6y + 9) \][/tex]
Now, combine like terms:
[tex]\[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 + 6y + 9 \][/tex]
Since \(x^2\) and \(y^2\) appear on both sides, we can cancel them out:
[tex]\[ -2x + 1 + 4 - 4y = -4x + 4 + 6y + 9 \][/tex]
Combine like terms again:
[tex]\[ -2x - 4y + 5 = -4x + 6y + 13 \][/tex]
Move all terms involving \(x\) and \(y\) to one side of the equation:
[tex]\[ 2x - 4x + 6y + 4y = 13 - 5 \][/tex]
Simplify:
[tex]\[ 2x + 10y = 8 \][/tex]
Divide the entire equation by 10 to isolate \(y\):
[tex]\[ x + 5y = 4 \][/tex]
Solving for \(y\), we get:
[tex]\[ y = \frac{1}{5}x - \frac{4}{5} \][/tex]
So, the set of points \((x, y)\) that are equidistant from \((1, 2)\) and \((2, -3)\) lie on the line:
[tex]\[ y = 0.2x - 0.8 \][/tex]
This is the final solution.
We are given two points \((1, 2)\) and \((2, -3)\) and we need to find the set of all points \((x, y)\) in the plane which are at an equal distance from these two given points.
To start, let's use the distance formula. The distance between any two points \((x_1, y_1)\) and \((x_2, y_2)\) in the plane is given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let's apply this to our problem.
1. The distance from point \((x, y)\) to the point \((1, 2)\):
[tex]\[ d_1 = \sqrt{(x - 1)^2 + (y - 2)^2} \][/tex]
2. The distance from point \((x, y)\) to the point \((2, -3)\):
[tex]\[ d_2 = \sqrt{(x - 2)^2 + (y + 3)^2} \][/tex]
Since we are looking for the points that are equidistant to \((1, 2)\) and \((2, -3)\):
[tex]\[ \sqrt{(x - 1)^2 + (y - 2)^2} = \sqrt{(x - 2)^2 + (y + 3)^2} \][/tex]
To isolate \(y\), we first eliminate the square roots by squaring both sides:
[tex]\[ ((x - 1)^2 + (y - 2)^2) = ((x - 2)^2 + (y + 3)^2) \][/tex]
Now we expand both sides:
[tex]\[ (x - 1)^2 + (y - 2)^2 = (x - 2)^2 + (y + 3)^2 \][/tex]
Expanding each term:
[tex]\[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = (x^2 - 4x + 4) + (y^2 + 6y + 9) \][/tex]
Now, combine like terms:
[tex]\[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 + 6y + 9 \][/tex]
Since \(x^2\) and \(y^2\) appear on both sides, we can cancel them out:
[tex]\[ -2x + 1 + 4 - 4y = -4x + 4 + 6y + 9 \][/tex]
Combine like terms again:
[tex]\[ -2x - 4y + 5 = -4x + 6y + 13 \][/tex]
Move all terms involving \(x\) and \(y\) to one side of the equation:
[tex]\[ 2x - 4x + 6y + 4y = 13 - 5 \][/tex]
Simplify:
[tex]\[ 2x + 10y = 8 \][/tex]
Divide the entire equation by 10 to isolate \(y\):
[tex]\[ x + 5y = 4 \][/tex]
Solving for \(y\), we get:
[tex]\[ y = \frac{1}{5}x - \frac{4}{5} \][/tex]
So, the set of points \((x, y)\) that are equidistant from \((1, 2)\) and \((2, -3)\) lie on the line:
[tex]\[ y = 0.2x - 0.8 \][/tex]
This is the final solution.
