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Which equation represents the line that is perpendicular to [tex]y=\frac{4}{5} x+23[/tex] and passes through [tex](-40,20)[/tex]?

A. [tex]y=-\frac{5}{4} x-15[/tex]
B. [tex]y=-\frac{5}{4} x-30[/tex]
C. [tex]y=\frac{4}{5} x+52[/tex]
D. [tex]y=\frac{4}{5} x-56[/tex]


Sagot :

To determine which equation represents the line that is perpendicular to [tex]\( y = \frac{4}{5}x + 23 \)[/tex] and passes through the point [tex]\((-40, 20)\)[/tex], follow these steps:

1. Determine the slope of the given line:
The equation of the given line is [tex]\( y = \frac{4}{5}x + 23 \)[/tex]. The slope [tex]\( m \)[/tex] of this line is [tex]\(\frac{4}{5}\)[/tex].

2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another is the negative reciprocal of the original slope. For the given slope [tex]\(\frac{4}{5}\)[/tex], the perpendicular slope [tex]\( m_{\perpendicular} \)[/tex] is:
[tex]\[ m_{\perpendicular} = -\frac{1}{m} = -\frac{1}{\frac{4}{5}} = -\frac{5}{4} \][/tex]

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

Given point [tex]\((-40, 20)\)[/tex] and slope [tex]\(-\frac{5}{4}\)[/tex], the equation becomes:
[tex]\[ y - 20 = -\frac{5}{4} (x + 40) \][/tex]

4. Simplify the equation to slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 20 = -\frac{5}{4} (x + 40) \][/tex]
Distribute the slope [tex]\(-\frac{5}{4}\)[/tex]:
[tex]\[ y - 20 = -\frac{5}{4}x - 50 \][/tex]
Add 20 to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{5}{4}x - 50 + 20 \][/tex]
Simplify the constants:
[tex]\[ y = -\frac{5}{4}x - 30 \][/tex]

Therefore, the equation of the line perpendicular to [tex]\( y = \frac{4}{5}x + 23 \)[/tex] and passing through [tex]\((-40, 20)\)[/tex] is [tex]\( y = -\frac{5}{4}x - 30 \)[/tex].

So, the correct answer is:
B. [tex]\( y = -\frac{5}{4}x - 30 \)[/tex]