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Consider the line [tex][tex]$y=7x-3$[/tex][/tex].

1. Find the equation of the line that is parallel to this line and passes through the point [tex][tex]$(2,-2)$[/tex][/tex].
2. Find the equation of the line that is perpendicular to this line and passes through the point [tex][tex]$(2,-2)$[/tex][/tex].

Equation of the parallel line: [tex]\square[/tex]

Equation of the perpendicular line: [tex]\square[/tex]

Sagot :

GinnyAnswer
To solve this problem, we need to find the equations of two lines related to the given line [tex]\( y = 7x - 3 \)[/tex]:

1. Equation of the line that is parallel to [tex]\( y = 7x - 3 \)[/tex] and passes through the point [tex]\( (2, -2) \)[/tex].
2. Equation of the line that is perpendicular to [tex]\( y = 7x - 3 \)[/tex] and passes through the point [tex]\( (2, -2) \)[/tex].

### Step-by-Step Solution

#### 1. Equation of the Parallel Line

Parallel lines have the same slope. Since the given line is [tex]\( y = 7x - 3 \)[/tex], its slope ([tex]\( m \)[/tex]) is 7.

We need the equation of a line with slope 7 that passes through the point [tex]\( (2, -2) \)[/tex]. Using the point-slope form of the equation of a line:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

Substitute [tex]\( m = 7 \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = -2 \)[/tex]:

[tex]\[ y - (-2) = 7(x - 2) \][/tex]
[tex]\[ y + 2 = 7(x - 2) \][/tex]

Expand and solve for [tex]\( y \)[/tex]:

[tex]\[ y + 2 = 7x - 14 \][/tex]
[tex]\[ y = 7x - 14 - 2 \][/tex]
[tex]\[ y = 7x - 16 \][/tex]

So, the equation of the parallel line is:

[tex]\[ \boxed{y = 7x - 16} \][/tex]

#### 2. Equation of the Perpendicular Line

Perpendicular lines have slopes that are negative reciprocals of each other. The slope of the given line is 7, so the slope of the perpendicular line is [tex]\( -\frac{1}{7} \)[/tex].

Using the point-slope form again with [tex]\( m = -\frac{1}{7} \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = -2 \)[/tex]:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

Substitute:

[tex]\[ y - (-2) = -\frac{1}{7}(x - 2) \][/tex]
[tex]\[ y + 2 = -\frac{1}{7}(x - 2) \][/tex]

Expand:

[tex]\[ y + 2 = -\frac{1}{7}x + \frac{2}{7} \][/tex]

Solve for [tex]\( y \)[/tex]:

[tex]\[ y = -\frac{1}{7}x + \frac{2}{7} - 2 \][/tex]
[tex]\[ y = -\frac{1}{7}x + \frac{2}{7} - \frac{14}{7} \][/tex]
[tex]\[ y = -\frac{1}{7}x - \frac{12}{7} \][/tex]

So, the equation of the perpendicular line is:

[tex]\[ \boxed{y = -\frac{1}{7}x - \frac{12}{7}} \][/tex]

Thus, we've found:
- Equation of the parallel line: [tex]\( y = 7x - 16 \)[/tex]
- Equation of the perpendicular line: [tex]\( y = -\frac{1}{7}x - \frac{12}{7} \)[/tex]
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