Let's solve both parts of the question step-by-step.
### Part 4: The angle between the lines [tex]\(2x + 3y = 4\)[/tex] and [tex]\(3x - 2y = 7\)[/tex]
First, we need to find the slopes ([tex]\(m\)[/tex]) of both lines:
1. For the line [tex]\(2x + 3y = 4\)[/tex]:
- Rewrite in slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[
3y = -2x + 4 \implies y = -\frac{2}{3}x + \frac{4}{3}
\][/tex]
- So, [tex]\(m_1 = -\frac{2}{3}\)[/tex].
2. For the line [tex]\(3x - 2y = 7\)[/tex]:
- Rewrite in slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[
-2y = -3x + 7 \implies y = \frac{3}{2}x - \frac{7}{2}
\][/tex]
- So, [tex]\(m_2 = \frac{3}{2}\)[/tex].
Next, we use the formula to find the tangent of the angle ([tex]\(\theta\)[/tex]) between two lines with slopes [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex]:
[tex]\[
\tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|
\][/tex]
Plugging in the values of [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex]:
[tex]\[
\tan(\theta) = \left|\frac{-\frac{2}{3} - \frac{3}{2}}{1 + \left(-\frac{2}{3}\right)\left(\frac{3}{2}\right)}\right|
\][/tex]
[tex]\[
= \left|\frac{-\frac{4}{6} - \frac{9}{6}}{1 + \left(-\frac{6}{6}\right)}\right|
\][/tex]
[tex]\[
= \left|\frac{-\frac{13}{6}}{1 - 1}\right|
\][/tex]
Note there is an error since the denominator became zero. This tells us that [tex]\(\theta = \frac{\pi}{2}\)[/tex] because the slopes imply perpendicular lines due to an undefined tangent value.
Therefore, the angle between the lines is:
[tex]\[
\boxed{\frac{\pi}{2}}
\][/tex]
### Part 5: The coordinates of the centroid of a triangle
The coordinates of the centroid [tex]\((G)\)[/tex] of a triangle whose vertices are [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] can be calculated using the formula:
[tex]\[
G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
\][/tex]
Therefore, the correct answer for the coordinates of the centroid is:
[tex]\[
\boxed{\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)}
\][/tex]
