Answer :
To determine the equation of the line that passes through the point [tex]\((8, -5)\)[/tex] and is parallel to one of the given lines, we need to follow these steps:
1. Identify the slopes of the given equations:
- The slope of [tex]\(y = \frac{3}{4} x + 1\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
- The slope of [tex]\(y = \frac{3}{4} x - 11\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
- The slope of [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] is [tex]\(-\frac{4}{3}\)[/tex].
- The slope of [tex]\(y = -\frac{4}{3} x - \frac{47}{3}\)[/tex] is [tex]\(-\frac{4}{3}\)[/tex].
2. Determine the equations of the lines through [tex]\((8, -5)\)[/tex] with the same slopes:
For lines with slope [tex]\(\frac{3}{4}\)[/tex]:
- Use the point-slope form: [tex]\(y = mx + b\)[/tex].
- Plug in the point [tex]\((8, -5)\)[/tex] and the slope [tex]\(\frac{3}{4}\)[/tex]:
[tex]\[ -5 = \frac{3}{4} \cdot 8 + b \][/tex]
- Solve for [tex]\(b\)[/tex]:
[tex]\[ -5 = 6 + b \implies b = -11 \][/tex]
- Therefore, the equation of the line is [tex]\(y = \frac{3}{4} x - 11\)[/tex].
For lines with slope [tex]\(-\frac{4}{3}\)[/tex]:
- Use the point-slope form: [tex]\(y = mx + b\)[/tex].
- Plug in the point [tex]\((8, -5)\)[/tex] and the slope [tex]\(-\frac{4}{3}\)[/tex]:
[tex]\[ -5 = -\frac{4}{3} \cdot 8 + b \][/tex]
- Solve for [tex]\(b\)[/tex]:
[tex]\[ -5 = -\frac{32}{3} + b \implies -5 + \frac{32}{3} = b \implies b = \frac{-15 + 32}{3} = \frac{17}{3} \][/tex]
- Therefore, the equation of the line is [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex].
3. Match these equations with the options provided:
- Option A [tex]\(y = \frac{3}{4} x + 1\)[/tex] does not match.
- Option B [tex]\(y = \frac{3}{4} x - 11\)[/tex] matches our first derived equation.
- Option C [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] matches our second derived equation.
- Option D [tex]\(y = -\frac{4}{3} x - \frac{47}{3}\)[/tex] does not match.
There are two correct answers for the given problem, which are:
- [tex]\(y = \frac{3}{4} x - 11\)[/tex] (Option B)
- [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] (Option C)
Either of these lines satisfies the conditions of passing through the point [tex]\((8, -5)\)[/tex] and being parallel to one of the given lines.
1. Identify the slopes of the given equations:
- The slope of [tex]\(y = \frac{3}{4} x + 1\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
- The slope of [tex]\(y = \frac{3}{4} x - 11\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
- The slope of [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] is [tex]\(-\frac{4}{3}\)[/tex].
- The slope of [tex]\(y = -\frac{4}{3} x - \frac{47}{3}\)[/tex] is [tex]\(-\frac{4}{3}\)[/tex].
2. Determine the equations of the lines through [tex]\((8, -5)\)[/tex] with the same slopes:
For lines with slope [tex]\(\frac{3}{4}\)[/tex]:
- Use the point-slope form: [tex]\(y = mx + b\)[/tex].
- Plug in the point [tex]\((8, -5)\)[/tex] and the slope [tex]\(\frac{3}{4}\)[/tex]:
[tex]\[ -5 = \frac{3}{4} \cdot 8 + b \][/tex]
- Solve for [tex]\(b\)[/tex]:
[tex]\[ -5 = 6 + b \implies b = -11 \][/tex]
- Therefore, the equation of the line is [tex]\(y = \frac{3}{4} x - 11\)[/tex].
For lines with slope [tex]\(-\frac{4}{3}\)[/tex]:
- Use the point-slope form: [tex]\(y = mx + b\)[/tex].
- Plug in the point [tex]\((8, -5)\)[/tex] and the slope [tex]\(-\frac{4}{3}\)[/tex]:
[tex]\[ -5 = -\frac{4}{3} \cdot 8 + b \][/tex]
- Solve for [tex]\(b\)[/tex]:
[tex]\[ -5 = -\frac{32}{3} + b \implies -5 + \frac{32}{3} = b \implies b = \frac{-15 + 32}{3} = \frac{17}{3} \][/tex]
- Therefore, the equation of the line is [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex].
3. Match these equations with the options provided:
- Option A [tex]\(y = \frac{3}{4} x + 1\)[/tex] does not match.
- Option B [tex]\(y = \frac{3}{4} x - 11\)[/tex] matches our first derived equation.
- Option C [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] matches our second derived equation.
- Option D [tex]\(y = -\frac{4}{3} x - \frac{47}{3}\)[/tex] does not match.
There are two correct answers for the given problem, which are:
- [tex]\(y = \frac{3}{4} x - 11\)[/tex] (Option B)
- [tex]\(y = -\frac{4}{3} x + \frac{17}{3}\)[/tex] (Option C)
Either of these lines satisfies the conditions of passing through the point [tex]\((8, -5)\)[/tex] and being parallel to one of the given lines.
