What is an equation of the line that is perpendicular to [tex]$y - 3 = -4(x + 2)$[/tex] and passes through the point [tex]$(-5, 7)$[/tex]?

A. [tex]$y - 7 = \frac{1}{4}(x + 5)$[/tex]

B. [tex][tex]$y - 7 = -4(x + 5)$[/tex][/tex]

C. [tex]$y + 7 = 4(x - 5)$[/tex]

D. [tex]$y + 7 = -\frac{1}{4}(x - 5)$[/tex]



Answer :

To find the equation of the line that is perpendicular to the given line and passes through the specific point, follow these steps:

### Step 1: Identify the slope of the given line

The given equation of the line is in point-slope form:
[tex]\[ y - 3 = -4(x + 2) \][/tex]

From this, you can see that the slope ([tex]\(m\)[/tex]) of the given line is [tex]\(-4\)[/tex].

### Step 2: Determine the slope of the perpendicular line

For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. If the slope of the given line is [tex]\(-4\)[/tex], then the slope of the line that is perpendicular to it, [tex]\(m_{\perpendicular}\)[/tex], will be the negative reciprocal of [tex]\(-4\)[/tex].

[tex]\[ m_{\perpendicular} = \frac{1}{-(-4)} = \frac{1}{4} \][/tex]

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

Now that we have the slope of the perpendicular line ([tex]\(\frac{1}{4}\)[/tex]) and a point it passes through ([tex]\(-5, 7\)[/tex]), we can use the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]

Here, [tex]\((x_1, y_1)\)[/tex] is [tex]\((-5, 7)\)[/tex] and [tex]\(m\)[/tex] is [tex]\(\frac{1}{4}\)[/tex].

### Step 4: Substitute the given point and slope into the point-slope formula

Substitute [tex]\((x_1, y_1) = (-5, 7)\)[/tex] and [tex]\(m = \frac{1}{4}\)[/tex] into the formula:

[tex]\[ y - 7 = \frac{1}{4}(x + 5) \][/tex]

This is the equation of the line that is perpendicular to the given line and passes through the point [tex]\((-5, 7)\)[/tex].

### Conclusion

The correct equation that represents the perpendicular line is:
[tex]\[ y - 7 = \frac{1}{4}(x + 5) \][/tex]

Therefore, the correct option is:

A. [tex]\(y-7=\frac{1}{4}(x+5)\)[/tex]