Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

The equation of the locus of the points equidistant from the points [tex]A (-2, 3)[/tex] and [tex]B (6, -5)[/tex] is:

1. [tex]x + y = 3[/tex]
2. [tex]x - y = 3[/tex]
3. [tex]2x + y = 3[/tex]
4. [tex]2x - y = 3[/tex]


Sagot :

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].