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Identify an equation in point-slope form for the line perpendicular to [tex][tex]$y=\frac{1}{4} x-7$[/tex][/tex] that passes through [tex][tex]$(-2,-6)$[/tex][/tex].

A. [tex][tex]$y+6=-4(x+2)$[/tex][/tex]

B. [tex][tex]$y+6=-\frac{1}{4}(x+2)$[/tex][/tex]

C. [tex][tex]$y+2=-4(x+6)$[/tex][/tex]

D. [tex][tex]$y-6=\frac{1}{4}(x-2)$[/tex][/tex]

Sagot :

GinnyAnswer
Certainly! Let's identify the equation in the point-slope form for the line perpendicular to [tex]\( y = \frac{1}{4} x - 7 \)[/tex] that passes through the point [tex]\((-2, -6)\)[/tex].

1. Determine the slope of the given line:
The given line is [tex]\( y = \frac{1}{4} x - 7 \)[/tex]. The slope (m) of this line is [tex]\( \frac{1}{4} \)[/tex].

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. For the original slope [tex]\( \frac{1}{4} \)[/tex], the negative reciprocal is:
[tex]\[ -\frac{1}{ \left( \frac{1}{4} \right) } = -4 \][/tex]

3. Use the point-slope form of the equation:
The point-slope form of a line's equation is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line. We have [tex]\((-2, -6)\)[/tex] as the point and [tex]\(-4\)[/tex] as the slope. Substituting these into the point-slope form:
[tex]\[ y - (-6) = -4(x - (-2)) \][/tex]
Simplifying:
[tex]\[ y + 6 = -4(x + 2) \][/tex]

Thus, the equation in point-slope form for the line perpendicular to [tex]\( y = \frac{1}{4} x - 7 \)[/tex] that passes through the point [tex]\((-2, -6)\)[/tex] is:

[tex]\[ y + 6 = -4(x + 2) \][/tex]

This corresponds to option A:
[tex]\[ \boxed{y + 6 = -4(x + 2)} \][/tex]
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