Direction: For the next two (2) items that follow:
The centroid of the triangle is at which one of the following points?
Finding the Centroid of a Triangle Defined by Lines
The problem asks us to find the centroid of the triangle formed by the intersection of three given lines: \(y = 3x\), \(y = 6x\), and \(y = 9\).
The centroid of a triangle is the point where the three medians of the triangle intersect. A median connects a vertex to the midpoint of the opposite side. To find the centroid, we first need to determine the coordinates of the three vertices of the triangle.
Identifying the Vertices of the Triangle
The vertices of the triangle are the points where the given lines intersect pairwise. We need to find the intersection points of:
- Line 1 (\(y = 3x\)) and Line 3 (\(y = 9\))
- Line 2 (\(y = 6x\)) and Line 3 (\(y = 9\))
- Line 1 (\(y = 3x\)) and Line 2 (\(y = 6x\))
Calculating the Intersection Points (Vertices)
- Intersection of \(y = 3x\) and \(y = 9\):
Substitute \(y = 9\) into the equation \(y = 3x\):
\(9 = 3x\)
Divide by 3 to solve for \(x\):
\(x = \frac{9}{3} = 3\)
So, the first vertex is \((3, 9)\).
- Intersection of \(y = 6x\) and \(y = 9\):
Substitute \(y = 9\) into the equation \(y = 6x\):
\(9 = 6x\)
Divide by 6 to solve for \(x\):
\(x = \frac{9}{6} = \frac{3}{2}\)
So, the second vertex is \(\left(\frac{3}{2}, 9\right)\).
- Intersection of \(y = 3x\) and \(y = 6x\):
Set the expressions for \(y\) equal to each other:
\(3x = 6x\)
Subtract \(3x\) from both sides:
\(0 = 6x - 3x\)
\(0 = 3x\)
Divide by 3:
\(x = 0\)
Substitute \(x = 0\) into either equation (e.g., \(y = 3x\)):
\(y = 3(0) = 0\)
So, the third vertex is \((0, 0)\).
The vertices of the triangle are \((3, 9)\), \(\left(\frac{3}{2}, 9\right)\), and \((0, 0)\).
Applying the Centroid Formula
Let the vertices be \((x_1, y_1) = (3, 9)\), \((x_2, y_2) = \left(\frac{3}{2}, 9\right)\), and \((x_3, y_3) = (0, 0)\).
The coordinates \((x_c, y_c)\) of the centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are given by the formula:
\(x_c = \frac{x_1 + x_2 + x_3}{3}\)
\(y_c = \frac{y_1 + y_2 + y_3}{3}\)
Now, we substitute the coordinates of our vertices into the formula:
Calculating the x-coordinate of the centroid:
\(x_c = \frac{3 + \frac{3}{2} + 0}{3}\)
To add the numbers in the numerator, find a common denominator:
\(3 + \frac{3}{2} + 0 = \frac{6}{2} + \frac{3}{2} + 0 = \frac{6 + 3}{2} = \frac{9}{2}\)
Now substitute this back into the \(x_c\) formula:
\(x_c = \frac{\frac{9}{2}}{3}\)
Dividing by 3 is the same as multiplying by \(\frac{1}{3}\):
\(x_c = \frac{9}{2} \times \frac{1}{3} = \frac{9}{6} = \frac{3}{2}\)
Calculating the y-coordinate of the centroid:
\(y_c = \frac{9 + 9 + 0}{3}\)
\(y_c = \frac{18}{3}\)
\(y_c = 6\)
Thus, the centroid of the triangle is at the point \(\left(\frac{3}{2}, 6\right)\).
Summary of Centroid Calculation
| Vertex | x-coordinate | y-coordinate |
|---|---|---|
| A | 3 | 9 |
| B | \(\frac{3}{2}\) | 9 |
| C | 0 | 0 |
| Centroid Coordinate | Calculation | Result |
|---|---|---|
| \(x_c\) | \(\frac{3 + \frac{3}{2} + 0}{3} = \frac{\frac{9}{2}}{3} = \frac{3}{2}\) | \(\frac{3}{2}\) |
| \(y_c\) | \(\frac{9 + 9 + 0}{3} = \frac{18}{3} = 6\) | 6 |
The centroid is located at \(\left(\frac{3}{2}, 6\right)\).
Revision Table: Key Concepts for Centroid and Triangle from Lines
| Concept | Description | How it applies here |
|---|---|---|
| Lines | Equations representing straight lines in a coordinate plane. | Given as \(y=3x\), \(y=6x\), \(y=9\). |
| Triangle | A polygon with three vertices and three sides. | Formed by the intersection points of the given lines. |
| Vertices | The points where the sides of a polygon meet. | Found by calculating the intersection points of the three lines. |
| Intersection Point | A point where two lines cross, satisfying both their equations. | Calculated by solving pairs of linear equations simultaneously. |
| Centroid | The geometric center of a triangle; the intersection of its medians. | Calculated using the average of the x-coordinates and y-coordinates of the vertices. |
| Centroid Formula | For vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), centroid is \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\). | Used to find the final centroid coordinates \(\left(\frac{3}{2}, 6\right)\). |
Additional Information: Centroid Properties and Medians
The centroid is always located inside the triangle. It has a unique property: it divides each median in a 2:1 ratio, with the longer segment being between the vertex and the centroid.
To find the median from a vertex, you would first find the midpoint of the opposite side using the midpoint formula \(\left(\frac{x_a+x_b}{2}, \frac{y_a+y_b}{2}\right)\). Then, the median is the line segment connecting the vertex to this midpoint.
For example, the midpoint of the side connecting \(\left(\frac{3}{2}, 9\right)\) and \((0, 0)\) is \(\left(\frac{\frac{3}{2} + 0}{2}, \frac{9 + 0}{2}\right) = \left(\frac{3}{4}, \frac{9}{2}\right)\). The median from vertex \((3, 9)\) connects \((3, 9)\) to \(\left(\frac{3}{4}, \frac{9}{2}\right)\). The centroid \(\left(\frac{3}{2}, 6\right)\) lies on this median.
Understanding how to find intersection points of lines and knowing the centroid formula are fundamental skills in coordinate geometry.
Important Questions from Centroid
If the centroid of a triangle formed by (7, x), (y, -6) and (9, 10) is (6, 3), then the values of x and y are respectively
What is the area of the triangle formed by these lines?
If \(\vec a, \vec b, \vec c\) , are the position vectors of the vertices A, B, C respectively of a triangle ABC and G is the centroid of the triangle, then what is \(\overrightarrow{AG}\) equal to ? .
The centroid of the triangle with vertices A(2, -3, 3), B(5, -3, -4) and C(2, -3, -2) is the point
If a vertex of a triangle is (1, 1) and the midpoints of two sides of the triangle through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
