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Sagot :
To find the equations of the lines perpendicular to the given lines and passing through the point [tex]\((2, 6)\)[/tex], follow these steps:
1. Given Line [tex]\(x = 2\)[/tex]:
- The line equation [tex]\(x = 2\)[/tex] is a vertical line.
- A line perpendicular to a vertical line must be horizontal.
- The equation of a horizontal line is of the form [tex]\(y = \text{constant}\)[/tex].
- Since this line passes through the point [tex]\((2, 6)\)[/tex], the constant must be the [tex]\(y\)[/tex]-coordinate of the point, which is 6.
- Therefore, the equation of the line perpendicular to [tex]\(x = 2\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(y = 6\)[/tex].
2. Given Line [tex]\(x = 6\)[/tex]:
- Similarly, the line [tex]\(x = 6\)[/tex] is also a vertical line.
- A line perpendicular to a vertical line is horizontal.
- Again, the equation of a horizontal line is of the form [tex]\(y = \text{constant}\)[/tex].
- Since this line must pass through the same point, the constant remains 6.
- Thus, the equation of the line perpendicular to [tex]\(x = 6\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(y = 6\)[/tex].
3. Given Line [tex]\(y = 2\)[/tex]:
- The line equation [tex]\(y = 2\)[/tex] is a horizontal line.
- A line perpendicular to a horizontal line must be vertical.
- The equation of a vertical line is of the form [tex]\(x = \text{constant}\)[/tex].
- Since this line passes through the point [tex]\((2, 6)\)[/tex], the constant must be the [tex]\(x\)[/tex]-coordinate of the point, which is 2.
- Therefore, the equation of the line perpendicular to [tex]\(y = 2\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(x = 2\)[/tex].
4. Given Line [tex]\(y = 6\)[/tex]:
- Similarly, the line [tex]\(y = 6\)[/tex] is a horizontal line.
- A line perpendicular to a horizontal line is vertical.
- Again, the equation of a vertical line is of the form [tex]\(x = \text{constant}\)[/tex].
- Since this line must pass through the point [tex]\((2, 6)\)[/tex], the constant remains 2.
- Thus, the equation of the line perpendicular to [tex]\(y = 6\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(x = 2\)[/tex].
Summarizing all the results, the lines perpendicular to the given lines and passing through the point [tex]\((2, 6)\)[/tex] are [tex]\(y = 6\)[/tex] and [tex]\(x = 2\)[/tex].
1. Given Line [tex]\(x = 2\)[/tex]:
- The line equation [tex]\(x = 2\)[/tex] is a vertical line.
- A line perpendicular to a vertical line must be horizontal.
- The equation of a horizontal line is of the form [tex]\(y = \text{constant}\)[/tex].
- Since this line passes through the point [tex]\((2, 6)\)[/tex], the constant must be the [tex]\(y\)[/tex]-coordinate of the point, which is 6.
- Therefore, the equation of the line perpendicular to [tex]\(x = 2\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(y = 6\)[/tex].
2. Given Line [tex]\(x = 6\)[/tex]:
- Similarly, the line [tex]\(x = 6\)[/tex] is also a vertical line.
- A line perpendicular to a vertical line is horizontal.
- Again, the equation of a horizontal line is of the form [tex]\(y = \text{constant}\)[/tex].
- Since this line must pass through the same point, the constant remains 6.
- Thus, the equation of the line perpendicular to [tex]\(x = 6\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(y = 6\)[/tex].
3. Given Line [tex]\(y = 2\)[/tex]:
- The line equation [tex]\(y = 2\)[/tex] is a horizontal line.
- A line perpendicular to a horizontal line must be vertical.
- The equation of a vertical line is of the form [tex]\(x = \text{constant}\)[/tex].
- Since this line passes through the point [tex]\((2, 6)\)[/tex], the constant must be the [tex]\(x\)[/tex]-coordinate of the point, which is 2.
- Therefore, the equation of the line perpendicular to [tex]\(y = 2\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(x = 2\)[/tex].
4. Given Line [tex]\(y = 6\)[/tex]:
- Similarly, the line [tex]\(y = 6\)[/tex] is a horizontal line.
- A line perpendicular to a horizontal line is vertical.
- Again, the equation of a vertical line is of the form [tex]\(x = \text{constant}\)[/tex].
- Since this line must pass through the point [tex]\((2, 6)\)[/tex], the constant remains 2.
- Thus, the equation of the line perpendicular to [tex]\(y = 6\)[/tex] and passing through [tex]\((2, 6)\)[/tex] is [tex]\(x = 2\)[/tex].
Summarizing all the results, the lines perpendicular to the given lines and passing through the point [tex]\((2, 6)\)[/tex] are [tex]\(y = 6\)[/tex] and [tex]\(x = 2\)[/tex].
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