Answer :
To find the equation of the locus of points that are equidistant from points \( A(-2, 3) \) and \( B(6, -5) \), we need to determine the perpendicular bisector of the line segment joining these two points.
### Step-by-Step Solution:
1. Finding the midpoint:
The midpoint \((M)\) of the line segment joining points \( A \) and \( B \) can be found using the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting the coordinates of \( A \) and \( B \):
[tex]\[ M = \left( \frac{-2 + 6}{2}, \frac{3 - 5}{2} \right) = \left( \frac{4}{2}, \frac{-2}{2} \right) = (2, -1) \][/tex]
So, the midpoint \( M \) is \( (2, -1) \).
2. Finding the slope of \( AB \):
The slope \( m_{AB} \) of the line segment \( AB \) is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates of \( A \) and \( B \):
[tex]\[ m_{AB} = \frac{-5 - 3}{6 - (-2)} = \frac{-8}{6 + 2} = \frac{-8}{8} = -1 \][/tex]
So, the slope \( m_{AB} \) is \( -1 \).
3. Finding the slope of the perpendicular bisector:
The slope of the perpendicular bisector is the negative reciprocal of the slope of \( AB \). Hence:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1 \][/tex]
So, the slope of the perpendicular bisector is \( 1 \).
4. Writing the equation of the perpendicular bisector:
The perpendicular bisector passes through the midpoint \( (2, -1) \) and has a slope of \( 1 \). Using the point-slope form of a line equation \( y - y_1 = m(x - x_1) \):
[tex]\[ y - (-1) = 1(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 1 = x - 2 \][/tex]
Rearranging to get the standard form:
[tex]\[ x - y = 3 \][/tex]
Therefore, the equation of the locus of points equidistant from \( A \) and \( B \) is:
[tex]\[ \boxed{x - y = 3} \][/tex]
So, the correct option is [tex]\( 2) \ x - y = 3 \)[/tex].
### Step-by-Step Solution:
1. Finding the midpoint:
The midpoint \((M)\) of the line segment joining points \( A \) and \( B \) can be found using the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting the coordinates of \( A \) and \( B \):
[tex]\[ M = \left( \frac{-2 + 6}{2}, \frac{3 - 5}{2} \right) = \left( \frac{4}{2}, \frac{-2}{2} \right) = (2, -1) \][/tex]
So, the midpoint \( M \) is \( (2, -1) \).
2. Finding the slope of \( AB \):
The slope \( m_{AB} \) of the line segment \( AB \) is given by:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the coordinates of \( A \) and \( B \):
[tex]\[ m_{AB} = \frac{-5 - 3}{6 - (-2)} = \frac{-8}{6 + 2} = \frac{-8}{8} = -1 \][/tex]
So, the slope \( m_{AB} \) is \( -1 \).
3. Finding the slope of the perpendicular bisector:
The slope of the perpendicular bisector is the negative reciprocal of the slope of \( AB \). Hence:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1 \][/tex]
So, the slope of the perpendicular bisector is \( 1 \).
4. Writing the equation of the perpendicular bisector:
The perpendicular bisector passes through the midpoint \( (2, -1) \) and has a slope of \( 1 \). Using the point-slope form of a line equation \( y - y_1 = m(x - x_1) \):
[tex]\[ y - (-1) = 1(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 1 = x - 2 \][/tex]
Rearranging to get the standard form:
[tex]\[ x - y = 3 \][/tex]
Therefore, the equation of the locus of points equidistant from \( A \) and \( B \) is:
[tex]\[ \boxed{x - y = 3} \][/tex]
So, the correct option is [tex]\( 2) \ x - y = 3 \)[/tex].
