the vertices of triangle ABC are A(-2,3), B(0,-3), and C(4,1). prove, by means of coordinate geometry, that the median to side BC is also the altitude to side BC.

To prove that the median to side BC is also the altitude to side BC, we need to show that the line segment connecting the midpoint of BC to vertex A is perpendicular to side BC.

Let's start by finding the midpoint of BC:
The coordinates of B are (0, -3) and the coordinates of C are (4, 1). The midpoint formula is given by the coordinates:
Midpoint of BC = [(x1 + x2)/2, (y1 + y2)/2]
Midpoint of BC = [(0 + 4) / 2, (-3 + 1) / 2]
Midpoint of BC = [2, -1]

Now, let's find the slope of the line passing through A(-2, 3) and the midpoint of BC (2, -1):
The slope formula is given by:
m = (y2 - y1)/(x2 - x1)

m = (-1 - 3) / (2 - (-2))
m = (-1 - 3) / (2 + 2)
m = -4 / 4
m = -1

Now, we need to find the slope of side BC, which is the line connecting points B(0, -3) and C(4, 1):
m' = (y2 - y1)/(x2 - x1)

m' = (1 - (-3)) / (4 - 0)
m' = (1 + 3) / 4
m' = 4 / 4
m' = 1

If the median to side BC is also the altitude to side BC, the product of their slopes should be -1. Let's check if this is true:

m * m' = -1 * 1 = -1

Since the product of the slopes is -1, we can conclude that the median to side BC is perpendicular to side BC, which means it is also the altitude to side BC.

To prove that the median to side BC is also the altitude to side BC, we need to show that the line segment connecting the midpoint of BC to vertex A is perpendicular to side BC.

Step 1: Find the midpoint of side BC
The midpoint (M) of side BC is calculated by taking the average of the x-coordinates and the average of the y-coordinates of B and C.

Coordinates of B: (0, -3)
Coordinates of C: (4, 1)

x-coordinate of M = (0 + 4) / 2 = 2
y-coordinate of M = (-3 + 1) / 2 = -1

So, midpoint M of side BC is (2, -1).

Step 2: Find the equation of the line passing through A and M
The equation of a line can be determined using the slope-intercept form, y = mx + b, where m is the slope of the line and b is the y-intercept.

Slope (m) = (y2 - y1) / (x2 - x1)
Coordinates of A: (-2, 3)
Coordinates of M: (2, -1)

m = (-1 - 3) / (2 - (-2)) = -4 / 4 = -1

Using the point-slope form, y - y1 = m(x - x1), we can substitute the values of A and M to get the equation of the line passing through them:

y - 3 = -1(x + 2)
y - 3 = -x - 2
y = -x + 1

So, the equation of the line passing through A and M is y = -x + 1.

Step 3: Find the equation of side BC
Using the coordinates of B and C, we can find the slope-intercept form of the line passing through them.

Slope (m) = (y2 - y1) / (x2 - x1)
Coordinates of B: (0, -3)
Coordinates of C: (4, 1)

m = (1 - (-3)) / (4 - 0) = 4 / 4 = 1

Using the point-slope form, y - y1 = m(x - x1), we can substitute the values of B to get the equation of side BC:

y - (-3) = 1(x - 0)
y + 3 = x
x - y = -3
x + (-y) + 3 = 0
x - y + 3 = 0

So, the equation of side BC is x - y + 3 = 0.

Step 4: Show that the lines are perpendicular
To prove that the median AM is perpendicular to side BC, we need to show that their slopes are negative reciprocals of each other.

The slope of the median AM is -1 (derived in Step 2).
The slope of the line BC is 1 (derived in Step 3).

Since the slopes of the two lines are negative reciprocals of each other (-1 * 1 = -1), it can be concluded that the median to side BC is also the altitude to side BC.

Therefore, the median to side BC is perpendicular to side BC, using coordinate geometry.

A median is a line joining the a vertex to the mid-point of the opposite side.

If the median is also an altitude to side BC, then the median should be perpendicular to BC.

Let D be the mid-point of BC. The coordinates of D should be rather obvious, being the mid-point of points B and C.

Compare the slopes of lines AC and lines BC. If the product of the two slopes equal -1, the lines are perpendicular, hence AD is an altitude to side BC.