Answer :
Let's solve the problem step-by-step:
### Part 1: Finding the Equation of the Diagonal AC
1. Identify and list the given points:
- Point A is [tex]\((2, 4)\)[/tex]
- Point C is [tex]\((4, 6)\)[/tex]
2. Calculate the midpoint of the line segment AC, which will also be the center of the square. The formula for the midpoint [tex]\( M \)[/tex] between points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Substituting in our given points:
[tex]\[ M = \left(\frac{2 + 4}{2}, \frac{4 + 6}{2}\right) = \left(3.0, 5.0\right) \][/tex]
The midpoint is [tex]\((3.0, 5.0)\)[/tex].
3. Determine the slope of AC. The slope [tex]\( m \)[/tex] between points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For our points A and C, the slope is:
[tex]\[ m_{AC} = \frac{6 - 4}{4 - 2} = \frac{2}{2} = 1 \][/tex]
4. Since the diagonals of a square intersect at right angles, the slope of the diagonal perpendicular to AC (let’s call it [tex]\( m \)[/tex]) can be found using the negative reciprocal of [tex]\( m_{AC} \)[/tex]:
[tex]\[ m = -\frac{1}{m_{AC}} = -\frac{1}{1} = -1 \][/tex]
5. Using the point-slope form of the equation of a line [tex]\( y - y_1 = m(x - x_1) \)[/tex] with point A(2, 4) and the calculated slope [tex]\(-1\)[/tex]:
[tex]\[ y - 4 = -1(x - 2) \][/tex]
Simplify it to the slope-intercept form [tex]\( y = mx + c \)[/tex]:
[tex]\[ y - 4 = -x + 2 \][/tex]
[tex]\[ y = -x + 2 + 4 \][/tex]
[tex]\[ y = -x + 6 \][/tex]
Thus, the equation of the diagonal is [tex]\( y = -x + 6 \)[/tex].
### Part 2: Find the Equation of the Median from Vertex B(10, 0) to the Origin O(0, 0)
1. The median from a vertex to the midpoint of the opposite side in a triangle is essentially the line joining the vertex to the midpoint of the opposite side.
Given points are:
- Point B is [tex]\((10, 0)\)[/tex]
- Point O is the origin, [tex]\((0, 0)\)[/tex]
2. Calculate the slope of the line segment BO using the slope formula:
[tex]\[ m_{BO} = \frac{0 - 0}{0 - 10} = 0 \][/tex]
3. Since the slope is [tex]\(0\)[/tex], the line segment is horizontal. The equation for a horizontal line passing through the origin is simply [tex]\(y = 0\)[/tex].
Therefore, the equation of the median from point B(10, 0) to the origin O(0, 0) is [tex]\(y = 0\)[/tex].
### Summary
- The midpoint of AC is [tex]\((3.0, 5.0)\)[/tex].
- The equation of the diagonal AC is [tex]\(y = -x + 6\)[/tex].
- The equation of the median from B(10, 0) to the origin O(0, 0) is [tex]\(y = 0\)[/tex].
### Part 1: Finding the Equation of the Diagonal AC
1. Identify and list the given points:
- Point A is [tex]\((2, 4)\)[/tex]
- Point C is [tex]\((4, 6)\)[/tex]
2. Calculate the midpoint of the line segment AC, which will also be the center of the square. The formula for the midpoint [tex]\( M \)[/tex] between points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Substituting in our given points:
[tex]\[ M = \left(\frac{2 + 4}{2}, \frac{4 + 6}{2}\right) = \left(3.0, 5.0\right) \][/tex]
The midpoint is [tex]\((3.0, 5.0)\)[/tex].
3. Determine the slope of AC. The slope [tex]\( m \)[/tex] between points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For our points A and C, the slope is:
[tex]\[ m_{AC} = \frac{6 - 4}{4 - 2} = \frac{2}{2} = 1 \][/tex]
4. Since the diagonals of a square intersect at right angles, the slope of the diagonal perpendicular to AC (let’s call it [tex]\( m \)[/tex]) can be found using the negative reciprocal of [tex]\( m_{AC} \)[/tex]:
[tex]\[ m = -\frac{1}{m_{AC}} = -\frac{1}{1} = -1 \][/tex]
5. Using the point-slope form of the equation of a line [tex]\( y - y_1 = m(x - x_1) \)[/tex] with point A(2, 4) and the calculated slope [tex]\(-1\)[/tex]:
[tex]\[ y - 4 = -1(x - 2) \][/tex]
Simplify it to the slope-intercept form [tex]\( y = mx + c \)[/tex]:
[tex]\[ y - 4 = -x + 2 \][/tex]
[tex]\[ y = -x + 2 + 4 \][/tex]
[tex]\[ y = -x + 6 \][/tex]
Thus, the equation of the diagonal is [tex]\( y = -x + 6 \)[/tex].
### Part 2: Find the Equation of the Median from Vertex B(10, 0) to the Origin O(0, 0)
1. The median from a vertex to the midpoint of the opposite side in a triangle is essentially the line joining the vertex to the midpoint of the opposite side.
Given points are:
- Point B is [tex]\((10, 0)\)[/tex]
- Point O is the origin, [tex]\((0, 0)\)[/tex]
2. Calculate the slope of the line segment BO using the slope formula:
[tex]\[ m_{BO} = \frac{0 - 0}{0 - 10} = 0 \][/tex]
3. Since the slope is [tex]\(0\)[/tex], the line segment is horizontal. The equation for a horizontal line passing through the origin is simply [tex]\(y = 0\)[/tex].
Therefore, the equation of the median from point B(10, 0) to the origin O(0, 0) is [tex]\(y = 0\)[/tex].
### Summary
- The midpoint of AC is [tex]\((3.0, 5.0)\)[/tex].
- The equation of the diagonal AC is [tex]\(y = -x + 6\)[/tex].
- The equation of the median from B(10, 0) to the origin O(0, 0) is [tex]\(y = 0\)[/tex].
