The problem asks to write the equation of a line that satisfies the following conditions:

1. It passes through point $C(-2, 3)$.
2. It is perpendicular to the line given by the equation $5x + 6y - 8 = 0$.

Solution:

1. Find the slope of the given line: The general form of a line is $Ax + By + C = 0$. For $5x + 6y - 8 = 0$, we can rewrite it in slope-intercept form: $6y = -5x + 8 \quad \Rightarrow \quad y = -\frac{5}{6}x + \frac{8}{6}.$ The slope of this line is: $m_1 = -\frac{5}{6}.$

2. Determine the slope of the perpendicular line: If two lines are perpendicular, the product of their slopes is $-1$. Thus, the slope $m_2$ of the perpendicular line satisfies: $m_1 \cdot m_2 = -1 \quad \Rightarrow \quad \left(-\frac{5}{6}\right) \cdot m_2 = -1 \quad \Rightarrow \quad m_2 = \frac{6}{5}.$

3. Write the equation of the line: The perpendicular line passes through $C(-2, 3)$ and has slope $m_2 = \frac{6}{5}$. Using the point-slope form of a line: $y - y_1 = m(x - x_1),$ where $(x_1, y_1) = (-2, 3)$, we substitute: $y - 3 = \frac{6}{5}(x + 2).$

4. Simplify the equation: Expand and rearrange: $y - 3 = \frac{6}{5}x + \frac{12}{5} \quad \Rightarrow \quad y = \frac{6}{5}x + \frac{12}{5} + 3.$ Convert $3$ to a fraction with a denominator of 5: $y = \frac{6}{5}x + \frac{12}{5} + \frac{15}{5} \quad \Rightarrow \quad y = \frac{6}{5}x + \frac{27}{5}.$

5. Final equation: The equation of the line is: $y = \frac{6}{5}x + \frac{27}{5}.$

Would you like a more detailed explanation for any step?

Related Questions:

1. What are the properties of perpendicular lines in terms of slopes?
2. How do you convert a general form equation to slope-intercept form?
3. Can you derive a line equation directly using two points instead of a slope and point?
4. What are other methods to find the perpendicular slope without directly solving for $m$?
5. How can this be visualized graphically?

Tip:

To confirm perpendicularity, always verify the negative reciprocal relationship between the slopes.