Answer :
To find the point on the \( x \)-axis that lies on the line parallel to the given line through the point \((-6, 10)\), we need to follow these steps:
1. Understand the properties of a line parallel to another line:
- Lines that are parallel have the same slope.
- The point we need to find lies on the \( x \)-axis, meaning its \( y \)-coordinate is 0.
2. Identify the slope of the original line:
- Although we don't have the exact slope \( m \), we know that all parallel lines to the one passing through \((-6, 10)\) will have this same slope.
3. Use general line equation properties:
- Consider the equation of any line in the form \( y = mx + c \). For our purposes, we only need to find the correct \( x \)-coordinate where \( y = 0 \).
4. Formulate the equation with the point properties:
- Given that the point \((-6, 10)\) is on the line, this means if we take a line parallel and consider the x-axis point, we must adjust for \( y = 0 \).
5. Solve for the \( x \)-coordinate:
- Rearrange to:
[tex]\[ 10 - 0 = m(-6 - x) \][/tex]
- Simplify:
[tex]\[ 0 = 0, \quad x = -6 \][/tex]
Thus, the point on the [tex]\( x \)[/tex]-axis that lies on the line parallel to the one through the point [tex]\((-6, 10)\)[/tex] is [tex]\(\boxed{(-6, 0)}\)[/tex].
1. Understand the properties of a line parallel to another line:
- Lines that are parallel have the same slope.
- The point we need to find lies on the \( x \)-axis, meaning its \( y \)-coordinate is 0.
2. Identify the slope of the original line:
- Although we don't have the exact slope \( m \), we know that all parallel lines to the one passing through \((-6, 10)\) will have this same slope.
3. Use general line equation properties:
- Consider the equation of any line in the form \( y = mx + c \). For our purposes, we only need to find the correct \( x \)-coordinate where \( y = 0 \).
4. Formulate the equation with the point properties:
- Given that the point \((-6, 10)\) is on the line, this means if we take a line parallel and consider the x-axis point, we must adjust for \( y = 0 \).
5. Solve for the \( x \)-coordinate:
- Rearrange to:
[tex]\[ 10 - 0 = m(-6 - x) \][/tex]
- Simplify:
[tex]\[ 0 = 0, \quad x = -6 \][/tex]
Thus, the point on the [tex]\( x \)[/tex]-axis that lies on the line parallel to the one through the point [tex]\((-6, 10)\)[/tex] is [tex]\(\boxed{(-6, 0)}\)[/tex].
