To find the equation of a line that passes through the point [tex]\( (2,5) \)[/tex] and is parallel to the line given by the equation [tex]\( x + 2y = 5 \)[/tex], follow these steps:
1. Determine the slope of the given line:
- Start by converting the given equation [tex]\( x + 2y = 5 \)[/tex] into slope-intercept form [tex]\( y = mx + b \)[/tex].
- Subtract [tex]\( x \)[/tex] from both sides: [tex]\( 2y = -x + 5 \)[/tex].
- Divide by 2: [tex]\( y = -\frac{1}{2}x + \frac{5}{2} \)[/tex].
So, the slope [tex]\( m \)[/tex] of the given line is [tex]\( -\frac{1}{2} \)[/tex].
2. Use the slope to find the equation of the parallel line:
- The slope of the line parallel to the given line will be the same, [tex]\( m = -\frac{1}{2} \)[/tex].
3. Form the equation using the point-slope form:
- The general form of a line using point-slope form is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is a point on the line.
- Use the point [tex]\( (2, 5) \)[/tex] to find the y-intercept [tex]\( b \)[/tex]. Substitute [tex]\( m = -\frac{1}{2} \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = 5 \)[/tex] into the equation [tex]\( y = mx + b \)[/tex]:
[tex]\[
5 = -\frac{1}{2}(2) + b
\][/tex]
[tex]\[
5 = -1 + b
\][/tex]
[tex]\[
b = 6
\][/tex]
4. Write the final equation:
- Substitute the slope [tex]\( m = -\frac{1}{2} \)[/tex] and the y-intercept [tex]\( b = 6 \)[/tex] into the slope-intercept form of the equation [tex]\( y = mx + b \)[/tex]:
[tex]\[
y = -\frac{1}{2} x + 6
\][/tex]
Hence, the equation of the line containing the point [tex]\( (2,5) \)[/tex] and parallel to the line [tex]\( x + 2y = 5 \)[/tex] is:
[tex]\[
y = -\frac{1}{2}x + 6
\][/tex]
