To find the equation of a straight line that contains the point (4, 1) and is perpendicular to the line [tex]\( y = 2x + 3 \)[/tex], follow these steps:
1. Determine the slope of the given line:
The given line is [tex]\( y = 2x + 3 \)[/tex]. From this equation, we can identify that the slope (m) of the line is 2.
2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope.
Therefore, the slope of the line perpendicular to [tex]\( y = 2x + 3 \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
3. Use the point-slope form of the equation of a line:
We know the slope of the perpendicular line [tex]\(-\frac{1}{2}\)[/tex] and it passes through the point (4, 1). The point-slope form of the equation of a line is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\( (x_1, y_1) = (4, 1) \)[/tex]. Substituting these values into the point-slope form:
[tex]\[
y - 1 = -\frac{1}{2}(x - 4)
\][/tex]
4. Simplify the equation to slope-intercept form [tex]\( y = mx + b \)[/tex]:
Expand and simplify the equation to get it into the form [tex]\( y = mx + b \)[/tex]:
[tex]\[
y - 1 = -\frac{1}{2}x + 2
\][/tex]
[tex]\[
y = -\frac{1}{2}x + 2 + 1
\][/tex]
[tex]\[
y = -\frac{1}{2}x + 3
\][/tex]
So, the equation of the line that is perpendicular to [tex]\( y = 2x + 3 \)[/tex] and passes through the point (4, 1) is:
[tex]\[
y = -\frac{1}{2}x + 3
\][/tex]
Thus, the slope of the perpendicular line is [tex]\(-0.5\)[/tex] and its equation in slope-intercept form is [tex]\( y = -\frac{1}{2}x + 3 \)[/tex].
