Question

What is the equation of a line that is perpendicular to -x+3y=9 and passes through the point (-3, 2)? Please enter your equation in the box.

Answer 1

The equation of the line perpendicular to (-x + 3y = 9) and passing through the point
`\((-3, 2)\) is \(3x + y = 7\)`.

Step-by-step explanation:

To determine the equation of a line perpendicular to (-x + 3y = 9), it is crucial to first express the given line in slope-intercept form (y = mx + b), where (m) is the slope. Rewriting (-x + 3y = 9) in this form yields (y =
`(1)/(3)x + 3\`), indicating a slope (
`\((1)/(3)\`). Since a line perpendicular to this one has a negative reciprocal slope, the perpendicular line's slope is (-3). Employing the point-slope form
`\((y - y_1) = m(x - x_1)\)` with the point (-3, 2), the equation becomes (3x + y = 7).

Understanding the concept of perpendicular lines involves recognizing that their slopes multiply to -1. In this case, the original line's slope
`(\((1)/(3)\))` and the perpendicular line's slope (-3) satisfy this condition, ensuring perpendicularity. The equation (3x + y = 7) is the result of applying these principles to find a line that meets the given criteria within the specified interval.

In conclusion, the process involves manipulating the given equation, identifying the slope, determining the negative reciprocal, and using the point-slope form to establish the equation of the perpendicular line. The final equation (3x + y = 7) satisfies these conditions and passes through the specified point.