Triangle PQR has vertices P(1,2), Q(25,2) and R(10,20).

Find the coordinates of the centroid.
Find the coordinates of the circumcenter.
Find the coordinates of the orthocenter.
Find the equation of the line.

1. P(1,2), Q(25,2), R(10,20).

X = (1+25+10) / 3 = 12.

Y = (2+2+20) / 3 = 8.

C(12,8). = Coordinates of Centroid.

2. SIDE PQ:
Xo = (1+25) / 2 = 13.

Yo = (2+2) / 2 = 2.

M(13,2).

m = (2-2) / (25-1) = 0/24 = 0.

m2 = -1/0 = Undefined = slope of perpendicular bisector(P.B.).

Eq1: X = 13 = Eq of P.B. of PQ = a ver line.

SIDE PR:

Xo = (1+10) / 2 = 5.5.

Yo = (2+20) / 2 = 11.

M(5.5,11).

m = (20-2) / (10-1) = 2.

m2 = -1/2 = slope of P.B. of PR.

Y = (-1/2)*5.5 + b = 11,
-2.75 + b = 11,
b = 11 + 2.75 = 13.75.

Eq2: Y = -X/2 + 13.75 = P.B. OF PR.

Substitute 13 for X:
Y = -13/2 + 13.75 = 7.25.

C(13,7.25).=Coordinates of circumcenter

To find the coordinates of the centroid, we can use the formula:

Centroid (G) = (x-coordinate of P + x-coordinate of Q + x-coordinate of R) / 3 , (y-coordinate of P + y-coordinate of Q + y-coordinate of R) / 3

Using the given coordinates, we can calculate the centroid as follows:

x-coordinate of G = (1 + 25 + 10) / 3 = 36 / 3 = 12
y-coordinate of G = (2 + 2 + 20) / 3 = 24 / 3 = 8

Therefore, the coordinates of the centroid (G) are G(12, 8).

To find the coordinates of the circumcenter, we can use the following equations:

Midpoint of PQ = ( (x-coordinate of P + x-coordinate of Q) / 2 , (y-coordinate of P + y-coordinate of Q) / 2 )
Midpoint of QR = ( (x-coordinate of Q + x-coordinate of R) / 2 , (y-coordinate of Q + y-coordinate of R) / 2 )

Using the given coordinates, we can calculate the midpoints as follows:

Midpoint of PQ = ( (1 + 25) / 2 , (2 + 2) / 2 ) = (13, 2)
Midpoint of QR = ( (25 + 10) / 2 , (2 + 20) / 2 ) = (17.5, 11)

To find the circumcenter (C), we can find the line perpendicular to PQ passing through the midpoint of PQ, and the line perpendicular to QR passing through the midpoint of QR. The intersection point of these two lines will give us the circumcenter.

Using the midpoint (13, 2) and the slope of PQ, we can find the equation of the line perpendicular to PQ passing through the midpoint of PQ.

Slope of PQ = (y-coordinate of Q - y-coordinate of P) / (x-coordinate of Q - x-coordinate of P) = (2 - 2) / (25 - 1) = 0 / 24 = 0

Since the slope of the line perpendicular to a line with slope 0 is undefined, the equation of the line perpendicular to PQ is x = 13.

Using the midpoint (17.5, 11) and the slope of QR, we can find the equation of the line perpendicular to QR passing through the midpoint of QR.

Slope of QR = (y-coordinate of R - y-coordinate of Q) / (x-coordinate of R - x-coordinate of Q) = (20 - 2) / (10 - 25) = 18 / (-15) = -1.2

The equation of the line perpendicular to QR passing through the midpoint of QR is given by: y - 11 = -1.2(x - 17.5)

Simplifying this equation, we get: y = -1.2x + 29

To find the coordinates of the circumcenter, we solve the system of equations formed by the two lines:
x = 13 and y = -1.2x + 29

Substituting x = 13 into the second equation, we get: y = -1.2(13) + 29 = -15.6 + 29 = 13.4

Therefore, the coordinates of the circumcenter (C) are C(13, 13.4).

To find the coordinates of the orthocenter, we can use the altitudes of the triangle.

The altitude of a triangle is a line that passes through a vertex and is perpendicular to the opposite side.

To find the equation of the altitude from vertex P, we need to find the equation of the line passing through point P and perpendicular to QR.

The slope of QR is -1.2, so the slope of the line perpendicular to QR is 1.2 (since the product of the slopes of perpendicular lines is -1).

Using point-slope form, the equation of the altitude from vertex P is:
y - 2 = 1.2(x - 1)

Simplifying this equation, we get: y = 1.2x - 0.8

Next, we find the equation of the altitude from vertex Q.
The slope of PQ is 0, so the slope of the line perpendicular to PQ is undefined.

Using point-slope form, the equation of the altitude from vertex Q is: x = 25

To find the coordinates of the orthocenter, we solve the system of equations formed by the two altitudes:
y = 1.2x - 0.8 and x = 25

Substituting x = 25 into the first equation, we get: y = 1.2(25) - 0.8 = 30 - 0.8 = 29.2

Therefore, the coordinates of the orthocenter (H) are H(25, 29.2).

To find the equation of the line, we can use the point-slope form of the equation:
y - y1 = m(x - x1)

where (x1, y1) is a point on the line and m is the slope of the line.

Using the coordinates (1, 2) and (25, 2), we can calculate the slope of the line:

Slope (m) = (y-coordinate of Q - y-coordinate of P) / (x-coordinate of Q - x-coordinate of P)
= (2 - 2) / (25 - 1)
= 0 / 24
= 0

Since the line is horizontal, the equation of the line is: y = 2

To find the coordinates of the centroid, we need to find the average of the x-coordinates and the average of the y-coordinates of the three vertices.

1. Calculate the average of the x-coordinates:
X-coordinate of centroid (C_x) = (x-coordinate of P + x-coordinate of Q + x-coordinate of R) / 3
C_x = (1 + 25 + 10) / 3 = 36 / 3 = 12

2. Calculate the average of the y-coordinates:
Y-coordinate of centroid (C_y) = (y-coordinate of P + y-coordinate of Q + y-coordinate of R) / 3
C_y = (2 + 2 + 20) / 3 = 24 / 3 = 8

Therefore, the coordinates of the centroid are (12, 8).

To find the coordinates of the circumcenter, we need to find the intersection point of the perpendicular bisectors of the sides of the triangle. Since there are no special properties in this case that make finding the circumcenter easy, we'll calculate it using equations.

1. Find the midpoint of each side:
Midpoint of PQ = [(x-coordinate of P + x-coordinate of Q) / 2, (y-coordinate of P + y-coordinate of Q) / 2]
= [(1 + 25) / 2, (2 + 2) / 2]
= [13, 2]

Midpoint of QR = [(x-coordinate of Q + x-coordinate of R) / 2, (y-coordinate of Q + y-coordinate of R) / 2]
= [(25 + 10) / 2, (2 + 20) / 2]
= [17.5, 11]

Midpoint of RP = [(x-coordinate of R + x-coordinate of P) / 2, (y-coordinate of R + y-coordinate of P) / 2]
= [(10 + 1) / 2, (20 + 2) / 2]
= [5.5, 11]

2. Find the slopes of the lines passing through each pair of midpoints:
Slope of PQ = (y-coordinate of Q - y-coordinate of P) / (x-coordinate of Q - x-coordinate of P)
= (2 - 2) / (25 - 1)
= 0 / 24
= 0

Slope of QR = (y-coordinate of R - y-coordinate of Q) / (x-coordinate of R - x-coordinate of Q)
= (20 - 2) / (10 - 25)
= 18 / (-15)
= -1.2

Slope of RP = (y-coordinate of P - y-coordinate of R) / (x-coordinate of P - x-coordinate of R)
= (2 - 20) / (1 - 10)
= -18 / (-9)
= 2

3. Find the equations of the perpendicular bisectors:
Equation of the perpendicular bisector of PQ:
Slope = -1 / slope of PQ = -1 / 0 = undefined
Since the slope is undefined, the perpendicular bisector is a vertical line passing through the midpoint of PQ, which has an x-coordinate of 13.
Therefore, the equation is x = 13.

Equation of the perpendicular bisector of QR:
Using the midpoint (17.5, 11) and the slope -1.2:
y - 11 = -1.2(x - 17.5)
y - 11 = -1.2x + 21
y = -1.2x + 32

Equation of the perpendicular bisector of RP:
Using the midpoint (5.5, 11) and the slope 2:
y - 11 = 2(x - 5.5)
y - 11 = 2x - 11
y = 2x

4. Solve the simultaneous equations of the perpendicular bisectors to find the circumcenter:
Solving x = 13 and y = -1.2x + 32, we get:
x = 13
y = -1.2(13) + 32 = 15.6

Solving x = 13 and y = 2x, we get:
x = 13
y = 2(13) = 26

Therefore, the circumcenter has coordinates (13, 15.6).

To find the coordinates of the orthocenter, we need to find the intersection point of the altitudes of the triangle. Similar to finding the circumcenter, this requires calculating equations.

1. Find the slopes of the lines passing through each side and perpendicular to the opposite vertex:
Slope of side PQ perpendicular to point R = -1 / slope of PQ = -1 / 0 = undefined
Slope of side QR perpendicular to point P = -1 / slope of QR = -1 / (-1.2) = 5 / 6
Slope of side RP perpendicular to point Q = -1 / slope of RP = -1 / 2 = -1/2

2. Use the point-slope form to find the equations of the altitudes:
Equation of the altitude from point R to side PQ:
Using the point (10,20) and the undefined slope:
x = 10

Equation of the altitude from point P to side QR:
Using the point (1,2) and the slope 5/6:
y - 2 = (5/6)(x - 1)
6y - 12 = 5x - 5
5x - 6y = -7

Equation of the altitude from point Q to side RP:
Using the point (25,2) and the slope -1/2:
y - 2 = (-1/2)(x - 25)
2y - 4 = -x + 25/2
x + 2y = 29/2

3. Solve the simultaneous equations of the altitudes to find the orthocenter:
Solving x = 10 and 5x - 6y = -7, we get:
x = 10
5(10) - 6y = -7
50 - 6y = -7
-6y = -57
y = 57/6 = 9.5

Solving x = 10 and x + 2y = 29/2, we get:
x = 10
10 + 2(9.5) = 29/2
10 + 19 = 29/2
29 = 29/2

Therefore, the orthocenter has coordinates (10, 9.5).

To find the equation of the line, we need two points on the line. Let's take the points P(1,2) and Q(25,2).

1. Find the slope of the line:
Slope = (y-coordinate of Q - y-coordinate of P) / (x-coordinate of Q - x-coordinate of P)
= (2 - 2) / (25 - 1)
= 0

Therefore, the slope of the line is 0.

2. Use the point-slope form of the equation to find the equation of the line:
y - y1 = m(x - x1)
Since the slope is 0, the equation simplifies to:
y - 2 = 0(x - 1)
y - 2 = 0
y = 2

Therefore, the equation of the line is y = 2.