To find the general form of the equation of the line that is perpendicular to the given line $x + 3y - 7 = 0$ and passes through the point $(3, -1)$, follow these steps:

Step 1: Find the slope of the given line

The given line equation is: $x + 3y - 7 = 0$ Rewrite this equation in the slope-intercept form $y = mx + b$ to find the slope: $3y = -x + 7$ $y = -\frac{1}{3}x + \frac{7}{3}$ Thus, the slope of the given line is $m_1 = -\frac{1}{3}$.

Step 2: Find the slope of the perpendicular line

The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. So, if the slope of the given line is $m_1 = -\frac{1}{3}$, then the slope of the perpendicular line, $m_2$, is: $m_2 = 3$

Step 3: Use the point-slope form of the line equation

The point-slope form of the equation of a line is: $y - y_1 = m(x - x_1)$ Here, $m = 3$, and the point $(x_1, y_1) = (3, -1)$.

Substitute these values into the point-slope form: $y - (-1) = 3(x - 3)$ $y + 1 = 3(x - 3)$ $y + 1 = 3x - 9$ $y = 3x - 10$

Step 4: Write the equation in general form

The general form of a line equation is $Ax + By + C = 0$. To convert $y = 3x - 10$ into this form: $3x - y - 10 = 0$

Thus, the general form of the equation of the line is: $3x - y - 10 = 0$

Final Answer:

The equation of the line perpendicular to $x + 3y - 7 = 0$ and passing through the point $(3, -1)$ is: $3x - y - 10 = 0$

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Would you like any further explanation or details on any of the steps?

Here are some related questions for further practice:

1. What is the slope of a line parallel to $x + 3y - 7 = 0$?
2. How do you find the equation of a line parallel to $x + 3y - 7 = 0$ and passing through $(4, 2)$?
3. What is the equation of a line perpendicular to $2x - 5y + 3 = 0$?
4. How do you find the intersection point of two lines?
5. What is the distance from the point $(3, -1)$ to the line $x + 3y - 7 = 0$?

Tip: The negative reciprocal of the slope of a line gives the slope of any line perpendicular to it.