Medians of a Triangle Practice

Question

Complete this assessment to review what you’ve learned. It will not count toward your grade.
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Question
Use the image to answer the question.

A scalene triangle made up of vertices upper A, upper B, and upper C, is plotted on a grid. Each of its vertices have their coordinates labeled, and they are connected to an interior point labeled upper D.

C = (4,7), A = (2,2), B = (8,2)

The three medians of △ABC
meet at a single point. What is the point of the centroid? Express all results in fractions.

(1 point)
centroid=
(
,
)

Answer 1

To find the centroid (the point where the three medians of a triangle meet), we can use the formula for the centroid of a triangle with vertices at coordinates \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

In the given triangle \(ABC\), the coordinates of the vertices are:

• \(A = (2, 2)\)
• \(B = (8, 2)\)
• \(C = (4, 7)\)

Now we can plug the coordinates into the centroid formula:

1. Calculate the x-coordinate of the centroid: \[ \text{x-coordinate} = \frac{2 + 8 + 4}{3} = \frac{14}{3} \]

2. Calculate the y-coordinate of the centroid: \[ \text{y-coordinate} = \frac{2 + 2 + 7}{3} = \frac{11}{3} \]

Thus, the centroid of triangle \(ABC\) is at:

\[ \text{Centroid} = \left( \frac{14}{3}, \frac{11}{3} \right) \]

So, the final answer is:

\[ \text{centroid} = \left( \frac{14}{3}, \frac{11}{3} \right) \]

Answer 2

i dont under stand the answer there are to many symbols remove them with out changing the answwr dont change the answer

Answer 3

Sure! Here’s the answer without the extra symbols:

centroid = (14/3, 11/3)

Answer 4

△ABC
is drawn on a coordinate plane with vertices A(1,3)
, B(6,6)
, and C(3,1)
and with medians indicating the midpoint of each of the lines AB
, BC
, and CA
. Prove that the medians meet at a single point by finding the centroid. Express all results in fractions.(1 point)
centroid=
(
,
)

Answer 5

To find the centroid of triangle ABC with vertices A(1, 3), B(6, 6), and C(3, 1), we can use the centroid formula:

\[ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Here, the vertices are:

• \(A = (1, 3)\)
• \(B = (6, 6)\)
• \(C = (3, 1)\)

Now let's compute the centroid:

1. Calculate the x-coordinate of the centroid: \[ \text{x-coordinate} = \frac{1 + 6 + 3}{3} = \frac{10}{3} \]

2. Calculate the y-coordinate of the centroid: \[ \text{y-coordinate} = \frac{3 + 6 + 1}{3} = \frac{10}{3} \]

Thus, the centroid of triangle ABC is at:

centroid = (10/3, 10/3)

Answer 6

Triangle XYZ
is drawn on a coordinate plane with vertices X(0,0)
, Y(3,6)
, and Z(4,3)
and with medians indicating the midpoint of each line XY
, YZ
, and ZX
. Prove that the medians meet at a single point by finding the centroid.(1 point)
centroid=
(
,
)

Answer 7

To find the centroid of triangle XYZ with vertices X(0, 0), Y(3, 6), and Z(4, 3), we will use the centroid formula:

\[ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Here, the vertices are:

• \(X = (0, 0)\)
• \(Y = (3, 6)\)
• \(Z = (4, 3)\)

Now let's compute the centroid:

1. Calculate the x-coordinate of the centroid: \[ \text{x-coordinate} = \frac{0 + 3 + 4}{3} = \frac{7}{3} \]

2. Calculate the y-coordinate of the centroid: \[ \text{y-coordinate} = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3 \]

Thus, the centroid of triangle XYZ is at:

centroid = (7/3, 3)

Answer 8

Medians of a Triangle Quick Check
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Question
Use the image to answer the question.

A triangle and 7 points are plotted on a coordinate plane. Median lines connect the vertices and median points intersecting at a point in the center of the triangle.

One way to prove that the medians of a triangle all meet at one point is by using arbitrary coordinates (0, 0)
, (a, 0)
, and (b, c)
to represent the three vertices of the triangle. Drag and drop the missing pieces of each step to complete the proof that all three medians intersect at point P.

(2 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Step 1: Draw the three medians of the triangle. Draw each median so that it starts at a Response area and ends at the Response area of the opposite side.

Step 2: Find the coordinates of each midpoint by taking the sum of the x-coordinates and dividing by 2 and taking the sum of the y-coordinates and dividing by 2.

The midpoint between (0, 0)
and (b, c)
is (b2, c2)
. The midpoint of (0, 0)
and (a, 0)
is (a2, 0)
. The midpoint of (a, 0)
and (b, c)
is (a+b2, c2)
.

Step 3: Find the coordinates of P, the point where the medians appear to intersect, on each median. Use the formula Response area.

Step 4: Find the coordinates of P on the median that starts at vertex (0, 0)
and ends at midpoint (a+b2, c2)
.

13(0, 0)+23(a+b2, c2)=(0, 0) + (a+b3, c3)=(a+b3, c3)
Find the coordinates of P on the median that starts at vertex Response area and ends at midpoint (b2, c2)
.

13(a, 0)+23(b2, c2)=(a3, 0)+(b3,c3)=(a+b3, c3)
Find the coordinates of P on the median that starts at vertex (b, c)
and ends at midpoint (a2, 0)
.

13(b, c)+23(a2, 0)=(b3,c3)+(a3,0)=(a+b3,c3)
Step 5: The coordinates of P on each median are Response area, which proves that the three medians of this generic triangle all intersect at the same point.

Answer 9

Let's fill in the missing pieces step by step for the proof that the medians of a triangle all meet at one point (point P).

Step 1:

Response to insert: starts at a vertex and ends at the midpoint of the opposite side.

Step 2:

1. Find the midpoints:
   • The midpoint between (0, 0) and (b, c) is \((\frac{b}{2}, \frac{c}{2})\).
   • The midpoint of (0, 0) and (a, 0) is \((\frac{a}{2}, 0)\).
   • The midpoint of (a, 0) and (b, c) is \((\frac{a+b}{2}, \frac{c}{2})\).

Step 3:

Response to insert: Use the formula for finding coordinates of points that divides the segment in a certain ratio, which is the section formula.

Step 4:

Find the coordinates of P on the median that starts at vertex (0, 0) and ends at midpoint (\(\frac{a+b}{2}, \frac{c}{2}\)): \[ \frac{1}{3}(0, 0) + \frac{2}{3}\left(\frac{a+b}{2}, \frac{c}{2}\right) = \left(0 + \frac{a+b}{3}, 0 + \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]

Find the coordinates of P on the median that starts at vertex (a, 0) and ends at midpoint (\(\frac{b}{2}, \frac{c}{2}\)): \[ \frac{1}{3}(a, 0) + \frac{2}{3}\left(\frac{b}{2}, \frac{c}{2}\right) = \left(\frac{a}{3}, 0\right) + \left(\frac{b}{3}, \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]

Find the coordinates of P on the median that starts at vertex (b, c) and ends at midpoint (\(\frac{a}{2}, 0\)): \[ \frac{1}{3}(b, c) + \frac{2}{3}\left(\frac{a}{2}, 0\right) = \left(\frac{b}{3}, \frac{c}{3}\right) + \left(\frac{a}{3}, 0\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]

Step 5:

Response to insert: The coordinates of P on each median are \((\frac{a+b}{3}, \frac{c}{3})\), which proves that the three medians of this generic triangle all intersect at the same point.

Summary of Responses:

1. starts at a vertex and ends at the midpoint of the opposite side.
2. \(\frac{b}{2}, \frac{c}{2}\); \(\frac{a}{2}, 0\); \(\frac{a+b}{2}, \frac{c}{2}\)
3. section formula
4. P = \((\frac{a+b}{3}, \frac{c}{3})\) for all medians
5. \((\frac{a+b}{3}, \frac{c}{3})\)

Feel free to ask if you need further assistance!