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What is an equation of the line that is perpendicular to [tex]y + 1 = -3(x - 5)[/tex] and passes through the point [tex](4, -6)[/tex]?

A. [tex]y + 6 = -3(x - 4)[/tex]

B. [tex]y - 6 = 3(x + 4)[/tex]

C. [tex]y - 6 = -\frac{1}{3}(x + 4)[/tex]

D. [tex]y + 6 = \frac{1}{3}(x - 4)[/tex]


Sagot :

To find the equation of a line that is perpendicular to the given line \( y + 1 = -3(x - 5) \) and passes through the point \((4, -6)\), let's follow these steps:

1. Determine the slope of the original line:
- The given equation \( y + 1 = -3(x - 5) \) can be rewritten in slope-intercept form \( y = mx + b \).
- Starting with \( y + 1 = -3(x - 5) \), distribute the \(-3\):
[tex]\[ y + 1 = -3x + 15 \][/tex]
- Isolate \( y \) to get:
[tex]\[ y = -3x + 14 \][/tex]
- From this, we see that the slope \( m \) of the original line is \(-3\).

2. Find the slope of the perpendicular line:
- The slope of a line perpendicular to another is the negative reciprocal of the original line's slope.
- The negative reciprocal of \(-3\) is \(\frac{1}{3}\). Hence, the slope \( m \) of the perpendicular line is \(\frac{1}{3}\).

3. Use the point-slope formula to find the equation:
- The point-slope form of a line's equation is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1) \) is a point on the line.
- Here, \( (x_1, y_1) = (4, -6) \) and \( m = \frac{1}{3} \).
- Substitute these values into the point-slope form:
[tex]\[ y - (-6) = \frac{1}{3}(x - 4) \][/tex]
- Simplify this equation:
[tex]\[ y + 6 = \frac{1}{3}(x - 4) \][/tex]

4. Select the correct answer choice:
- Comparing this final form with the given options, we see that it matches option D:
[tex]\[ y + 6 = \frac{1}{3}(x - 4) \][/tex]

Therefore, the equation of the line that is perpendicular to \( y + 1 = -3(x - 5) \) and passes through the point \((4, -6)\) is:
[tex]\[ \boxed{y + 6 = \frac{1}{3}(x - 4)} \][/tex]