Answer :
To find the equation of a straight line through the centroid of a triangle with given vertices [tex]\((3, -4)\)[/tex], [tex]\((-2, 1)\)[/tex], and [tex]\((5, 0)\)[/tex] and perpendicular to the line [tex]\(x - 3y = 4\)[/tex], we can follow these steps:
### Step 1: Calculate the Centroid of the Triangle
The centroid (G) of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \][/tex]
Given vertices:
[tex]\((x_1, y_1) = (3, -4)\)[/tex]
[tex]\((x_2, y_2) = (-2, 1)\)[/tex]
[tex]\((x_3, y_3) = (5, 0)\)[/tex]
Calculate the coordinates of the centroid:
[tex]\[ G_x = \frac{3 + (-2) + 5}{3} = \frac{6}{3} = 2 \][/tex]
[tex]\[ G_y = \frac{-4 + 1 + 0}{3} = \frac{-3}{3} = -1 \][/tex]
So the centroid [tex]\(G\)[/tex] is at:
[tex]\[ G = (2, -1) \][/tex]
### Step 2: Find the Slope of the Given Line
The given line equation is:
[tex]\[ x - 3y = 4 \][/tex]
To find the slope, we need to write it in slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ x - 3y = 4 \quad \Rightarrow \quad -3y = -x + 4 \quad \Rightarrow \quad y = \frac{1}{3}x - \frac{4}{3} \][/tex]
The slope [tex]\(m\)[/tex] of the given line is:
[tex]\[ m = \frac{1}{3} \][/tex]
### Step 3: Find the Slope of the Perpendicular Line
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the other line. Therefore, the slope [tex]\(m'\)[/tex] of the line perpendicular to [tex]\(x - 3y = 4\)[/tex] is:
[tex]\[ m' = -3 \][/tex]
### Step 4: Write the Equation of the Line Perpendicular Through the Centroid
We now know the slope of the line and a point through which it passes (the centroid [tex]\(G\)[/tex]). We can use the point-slope form of the equation of a line given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\( (x_1, y_1) = (2, -1) \)[/tex] and [tex]\(m = -3\)[/tex]:
[tex]\[ y - (-1) = -3(x - 2) \][/tex]
[tex]\[\Rightarrow y + 1 = -3(x - 2) \][/tex]
[tex]\[\Rightarrow y + 1 = -3x + 6 \][/tex]
[tex]\[\Rightarrow y = -3x + 5 \][/tex]
### Step 5: Convert to Standard Form
To convert [tex]\(y = -3x + 5\)[/tex] to standard form [tex]\(Ax + By = C\)[/tex]:
[tex]\[ y + 3x = 5 \][/tex]
[tex]\[\Rightarrow 3x + y = 5 \][/tex]
### Final Answer:
The equation of the line through the centroid of the triangle and perpendicular to the line [tex]\(x - 3y = 4\)[/tex] is:
[tex]\[ 3x + y = 5 \][/tex]
### Step 1: Calculate the Centroid of the Triangle
The centroid (G) of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \][/tex]
Given vertices:
[tex]\((x_1, y_1) = (3, -4)\)[/tex]
[tex]\((x_2, y_2) = (-2, 1)\)[/tex]
[tex]\((x_3, y_3) = (5, 0)\)[/tex]
Calculate the coordinates of the centroid:
[tex]\[ G_x = \frac{3 + (-2) + 5}{3} = \frac{6}{3} = 2 \][/tex]
[tex]\[ G_y = \frac{-4 + 1 + 0}{3} = \frac{-3}{3} = -1 \][/tex]
So the centroid [tex]\(G\)[/tex] is at:
[tex]\[ G = (2, -1) \][/tex]
### Step 2: Find the Slope of the Given Line
The given line equation is:
[tex]\[ x - 3y = 4 \][/tex]
To find the slope, we need to write it in slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ x - 3y = 4 \quad \Rightarrow \quad -3y = -x + 4 \quad \Rightarrow \quad y = \frac{1}{3}x - \frac{4}{3} \][/tex]
The slope [tex]\(m\)[/tex] of the given line is:
[tex]\[ m = \frac{1}{3} \][/tex]
### Step 3: Find the Slope of the Perpendicular Line
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the other line. Therefore, the slope [tex]\(m'\)[/tex] of the line perpendicular to [tex]\(x - 3y = 4\)[/tex] is:
[tex]\[ m' = -3 \][/tex]
### Step 4: Write the Equation of the Line Perpendicular Through the Centroid
We now know the slope of the line and a point through which it passes (the centroid [tex]\(G\)[/tex]). We can use the point-slope form of the equation of a line given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\( (x_1, y_1) = (2, -1) \)[/tex] and [tex]\(m = -3\)[/tex]:
[tex]\[ y - (-1) = -3(x - 2) \][/tex]
[tex]\[\Rightarrow y + 1 = -3(x - 2) \][/tex]
[tex]\[\Rightarrow y + 1 = -3x + 6 \][/tex]
[tex]\[\Rightarrow y = -3x + 5 \][/tex]
### Step 5: Convert to Standard Form
To convert [tex]\(y = -3x + 5\)[/tex] to standard form [tex]\(Ax + By = C\)[/tex]:
[tex]\[ y + 3x = 5 \][/tex]
[tex]\[\Rightarrow 3x + y = 5 \][/tex]
### Final Answer:
The equation of the line through the centroid of the triangle and perpendicular to the line [tex]\(x - 3y = 4\)[/tex] is:
[tex]\[ 3x + y = 5 \][/tex]
