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Consider the line [tex]$y=\frac{2}{7}x-6$[/tex].

1. Find the equation of the line that is parallel to this line and passes through the point [tex]$(4,3)$[/tex].
2. Find the equation of the line that is perpendicular to this line and passes through the point [tex]$(4,3)$[/tex].

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

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

Sagot :

GinnyAnswer
Let's analyze the problem step by step.

### Step 1: Equation of the given line
The given line equation is:
[tex]\[ y = \frac{2}{7}x - 6 \][/tex]
The slope of this line is \(\frac{2}{7}\).

### Step 2: Parallel Line through (4, 3)
For a line to be parallel to the given line, it must have the same slope. So, the slope of the parallel line will also be \(\frac{2}{7}\).

Using the point-slope form of the line equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where \((x_1, y_1)\) is the point the line passes through and \(m\) is the slope. Here, \((x_1, y_1) = (4, 3)\) and \(m = \frac{2}{7}\).

Substitute the values:
[tex]\[ y - 3 = \frac{2}{7}(x - 4) \][/tex]

Solving for \(y\):
[tex]\[ y - 3 = \frac{2}{7}x - \frac{8}{7} \][/tex]
[tex]\[ y = \frac{2}{7}x - \frac{8}{7} + 3 \][/tex]
[tex]\[ y = \frac{2}{7}x - \frac{8}{7} + \frac{21}{7} \][/tex]
[tex]\[ y = \frac{2}{7}x + \frac{13}{7} \][/tex]

So, the equation of the parallel line is:
[tex]\[ y = 0.285714285714286 x + 1.85714285714286 \][/tex]

### Step 3: Perpendicular Line through (4, 3)
For a line to be perpendicular to the given line, its slope must be the negative reciprocal of the slope of the given line. The slope of the given line is \(\frac{2}{7}\), so the slope of the perpendicular line will be:
[tex]\[ m_{\perpendicular} = -\frac{7}{2} \][/tex]

Using the same point-slope form:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substitute \((x_1, y_1) = (4, 3)\) and \(m = -\frac{7}{2}\):

[tex]\[ y - 3 = -\frac{7}{2}(x - 4) \][/tex]

Solving for \(y\):
[tex]\[ y - 3 = -\frac{7}{2}x + 14 \][/tex]
[tex]\[ y = -\frac{7}{2}x + 14 + 3 \][/tex]
[tex]\[ y = -\frac{7}{2}x + 17 \][/tex]

So, the equation of the perpendicular line is:
[tex]\[ y = 17 - 3.5 x \][/tex]

### Summary of Results

- Equation of the parallel line: [tex]\[ y = 0.285714285714286 x + 1.85714285714286 \][/tex]
- Equation of the perpendicular line: [tex]\[ y = 17 - 3.5 x \][/tex]

Feel free to fill these results into the squares provided in your question:

Equation of parallel line:

[tex]\[ y = 0.285714285714286 x + 1.85714285714286 \][/tex]

Equation of perpendicular line:

[tex]\[ y = 17 - 3.5 x \][/tex]
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