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Which equation represents the line that is perpendicular to [tex]\( y = \frac{3}{4}x + 1 \)[/tex] and passes through [tex]\((-5, 11)\)[/tex]?

A. [tex]\( y = -\frac{4}{3}x + \frac{13}{3} \)[/tex]

B. [tex]\( y = -\frac{4}{3}x + \frac{29}{3} \)[/tex]

C. [tex]\( y = \frac{3}{4}x + \frac{59}{4} \)[/tex]

D. [tex]\( y = \frac{3}{4}x - \frac{53}{4} \)[/tex]


Sagot :

To find the equation of the line that is perpendicular to the given line and passes through the point (-5, 11), follow these steps:

1. Identify the slope of the given line:
The equation of the given line is [tex]\( y = \frac{3}{4}x + 1 \)[/tex].
The slope of this line is [tex]\(\frac{3}{4}\)[/tex].

2. Determine 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. The negative reciprocal of [tex]\(\frac{3}{4}\)[/tex] is [tex]\(-\frac{4}{3}\)[/tex].
Therefore, the slope of the perpendicular line is [tex]\(-\frac{4}{3}\)[/tex].

3. Use the point-slope form of a line equation:
The point-slope form of a line equation is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( m \)[/tex] is the slope, and [tex]\((x_1, y_1)\)[/tex] is the point through which the line passes.

4. Substitute the slope and the given point:
Given the point [tex]\((-5, 11)\)[/tex] and the slope [tex]\(-\frac{4}{3}\)[/tex]:
[tex]\[ y - 11 = -\frac{4}{3}(x + 5) \][/tex]

5. Simplify the equation:
[tex]\[ y - 11 = -\frac{4}{3}(x + 5) \][/tex]
Distribute the [tex]\(-\frac{4}{3}\)[/tex]:
[tex]\[ y - 11 = -\frac{4}{3}x - \frac{4}{3} \times 5 \][/tex]
Simplify:
[tex]\[ y - 11 = -\frac{4}{3}x - \frac{20}{3} \][/tex]
Add 11 to both sides to isolate [tex]\(y\)[/tex]:
[tex]\[ y = -\frac{4}{3}x - \frac{20}{3} + 11 \][/tex]
Convert 11 to a fraction with a denominator of 3:
[tex]\[ 11 = \frac{33}{3} \][/tex]
Substitute back in:
[tex]\[ y = -\frac{4}{3}x - \frac{20}{3} + \frac{33}{3} \][/tex]
Combine the constant terms:
[tex]\[ y = -\frac{4}{3}x + \frac{13}{3} \][/tex]

Therefore, the equation of the line perpendicular to [tex]\( y = \frac{3}{4}x + 1 \)[/tex] and passing through the point [tex]\((-5, 11)\)[/tex] is:
[tex]\[ \boxed{y = -\frac{4}{3}x + \frac{13}{3}} \][/tex]

Thus, the correct answer is:
A. [tex]\( y = -\frac{4}{3}x + \frac{13}{3} \)[/tex]