What is the equation of the line that is perpendicular to and has the same [tex]y[/tex]-intercept as the given line?

Given line: [tex]y = \frac{1}{5} x + 1[/tex]

A. [tex]y = \frac{1}{5} x + 5[/tex]
B. [tex]y = 5 x + 1[/tex]
C. [tex]y = 5 x + 5[/tex]



Answer :

To find the equation of the line that is perpendicular to a given line and has the same [tex]\( y \)[/tex]-intercept as that given line, we need to follow these steps:

1. Identify the slope of the given line:

The given line is [tex]\( y = \frac{1}{5}x + 1 \)[/tex]. The slope-intercept form of a line is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept. Here, the slope [tex]\( m \)[/tex] of the given line is [tex]\( \frac{1}{5} \)[/tex].

2. Find the slope of the perpendicular line:

Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. If [tex]\( m_1 \)[/tex] is the slope of the given line, then the slope [tex]\( m_2 \)[/tex] of the perpendicular line is the negative reciprocal of [tex]\( m_1 \)[/tex].

For the given line, [tex]\( m_1 = \frac{1}{5} \)[/tex]. Therefore, the slope of the perpendicular line [tex]\( m_2 \)[/tex] is:
[tex]\[ m_2 = -\frac{1}{\left( \frac{1}{5} \right)} = -5 \][/tex]

3. Use the same [tex]\( y \)[/tex]-intercept as the given line:

The [tex]\( y \)[/tex]-intercept of the given line is [tex]\( 1 \)[/tex], since the given line is [tex]\( y = \frac{1}{5}x + 1 \)[/tex]. The new perpendicular line will share this same [tex]\( y \)[/tex]-intercept.

4. Write the equation of the perpendicular line:

Using the slope [tex]\( m_2 = -5 \)[/tex] and the [tex]\( y \)[/tex]-intercept [tex]\(1\)[/tex], the equation of the perpendicular line in slope-intercept form ([tex]\( y = mx + b \)[/tex]) is:
[tex]\[ y = -5x + 1 \][/tex]

Now, looking at the given options:
- [tex]\( y = \frac{1}{5}x + 1 \)[/tex]
- [tex]\( y = \frac{1}{5}x + 5 \)[/tex]
- [tex]\( y = 5x + 1 \)[/tex]
- [tex]\( y = 5x + 5 \)[/tex]

None of the options directly match the perpendicular line we derived, [tex]\( y = -5x + 1 \)[/tex]. Therefore, based on the correct calculations, the equation of the line that is perpendicular to and has the same [tex]\( y \)[/tex]-intercept as the given line should be derived as [tex]\( y = -5x + 1 \)[/tex].

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