Answer :
Let's tackle these problems one by one.
### (a) Its distance from the x-axis is always 8 units.
To find the equation of the locus of a point whose distance from the x-axis is always 8 units, we need to understand what this means. The distance of a point [tex]\((x, y)\)[/tex] from the x-axis is represented by the absolute value of its y-coordinate, [tex]\(|y|\)[/tex].
Given that this distance is always 8 units, we have:
[tex]\[ |y| = 8 \][/tex]
This equation tells us that [tex]\(y\)[/tex] can be either 8 or -8. Hence, the points that satisfy this condition will have their y-coordinates as [tex]\(8\)[/tex] or [tex]\(-8\)[/tex], regardless of their x-coordinates.
Thus, the equations representing the locus are:
[tex]\[ y = 8 \][/tex]
and
[tex]\[ y = -8 \][/tex]
So, the locus is represented by these two parallel lines y = 8 and y = -8.
### (b)
Without a specific query for part (b), it's challenging to provide a meaningful answer. If you meant another specific geometric condition for part (b), please provide it, and I'd be happy to help!
### (a) Its distance from the x-axis is always 8 units.
To find the equation of the locus of a point whose distance from the x-axis is always 8 units, we need to understand what this means. The distance of a point [tex]\((x, y)\)[/tex] from the x-axis is represented by the absolute value of its y-coordinate, [tex]\(|y|\)[/tex].
Given that this distance is always 8 units, we have:
[tex]\[ |y| = 8 \][/tex]
This equation tells us that [tex]\(y\)[/tex] can be either 8 or -8. Hence, the points that satisfy this condition will have their y-coordinates as [tex]\(8\)[/tex] or [tex]\(-8\)[/tex], regardless of their x-coordinates.
Thus, the equations representing the locus are:
[tex]\[ y = 8 \][/tex]
and
[tex]\[ y = -8 \][/tex]
So, the locus is represented by these two parallel lines y = 8 and y = -8.
### (b)
Without a specific query for part (b), it's challenging to provide a meaningful answer. If you meant another specific geometric condition for part (b), please provide it, and I'd be happy to help!