Answer :
We start by noting that parallel lines have the same slope. Since the line
[tex]$$
y = 2x + 3
$$[/tex]
has a slope of [tex]$2$[/tex], any line parallel to it will also have a slope of [tex]$2$[/tex].
Let the equation of the line [tex]$L_1$[/tex] be
[tex]$$
y = 2x + b.
$$[/tex]
Because [tex]$L_1$[/tex] passes through the point [tex]$P(2,6)$[/tex], we substitute [tex]$x = 2$[/tex] and [tex]$y = 6$[/tex] into the equation:
[tex]$$
6 = 2(2) + b.
$$[/tex]
Simplify to find [tex]$b$[/tex]:
[tex]$$
6 = 4 + b \quad \Longrightarrow \quad b = 2.
$$[/tex]
Thus, the equation of [tex]$L_1$[/tex] is:
[tex]$$
y = 2x + 2.
$$[/tex]
Next, we consider the line [tex]$L_2$[/tex], which is perpendicular to [tex]$L_1$[/tex] at the point [tex]$P(2,6)$[/tex]. For two lines to be perpendicular, the product of their slopes must be [tex]$-1$[/tex]. Since the slope of [tex]$L_1$[/tex] is [tex]$2$[/tex], the slope of [tex]$L_2$[/tex], denoted by [tex]$m_{L_2}$[/tex], is:
[tex]$$
m_{L_2} = -\frac{1}{2}.
$$[/tex]
Now, writing the equation of [tex]$L_2$[/tex] in point-slope form, we have:
[tex]$$
y - 6 = -\frac{1}{2}(x - 2).
$$[/tex]
To find the [tex]$y$[/tex]-intercept form, distribute and rearrange:
[tex]$$
y - 6 = -\frac{1}{2}x + 1.
$$[/tex]
Add [tex]$6$[/tex] to both sides:
[tex]$$
y = -\frac{1}{2}x + 7.
$$[/tex]
In summary, the detailed solutions are:
1. The equation of the line [tex]$L_1$[/tex] that is parallel to [tex]$y = 2x+3$[/tex] and passes through [tex]$P(2,6)$[/tex] is:
[tex]$$
y = 2x + 2.
$$[/tex]
2. The slope of the line [tex]$L_2$[/tex], which is perpendicular to [tex]$L_1$[/tex], is:
[tex]$$
-\frac{1}{2},
$$[/tex]
and its equation, using point [tex]$P(2,6)$[/tex], is:
[tex]$$
y = -\frac{1}{2}x + 7.
$$[/tex]
[tex]$$
y = 2x + 3
$$[/tex]
has a slope of [tex]$2$[/tex], any line parallel to it will also have a slope of [tex]$2$[/tex].
Let the equation of the line [tex]$L_1$[/tex] be
[tex]$$
y = 2x + b.
$$[/tex]
Because [tex]$L_1$[/tex] passes through the point [tex]$P(2,6)$[/tex], we substitute [tex]$x = 2$[/tex] and [tex]$y = 6$[/tex] into the equation:
[tex]$$
6 = 2(2) + b.
$$[/tex]
Simplify to find [tex]$b$[/tex]:
[tex]$$
6 = 4 + b \quad \Longrightarrow \quad b = 2.
$$[/tex]
Thus, the equation of [tex]$L_1$[/tex] is:
[tex]$$
y = 2x + 2.
$$[/tex]
Next, we consider the line [tex]$L_2$[/tex], which is perpendicular to [tex]$L_1$[/tex] at the point [tex]$P(2,6)$[/tex]. For two lines to be perpendicular, the product of their slopes must be [tex]$-1$[/tex]. Since the slope of [tex]$L_1$[/tex] is [tex]$2$[/tex], the slope of [tex]$L_2$[/tex], denoted by [tex]$m_{L_2}$[/tex], is:
[tex]$$
m_{L_2} = -\frac{1}{2}.
$$[/tex]
Now, writing the equation of [tex]$L_2$[/tex] in point-slope form, we have:
[tex]$$
y - 6 = -\frac{1}{2}(x - 2).
$$[/tex]
To find the [tex]$y$[/tex]-intercept form, distribute and rearrange:
[tex]$$
y - 6 = -\frac{1}{2}x + 1.
$$[/tex]
Add [tex]$6$[/tex] to both sides:
[tex]$$
y = -\frac{1}{2}x + 7.
$$[/tex]
In summary, the detailed solutions are:
1. The equation of the line [tex]$L_1$[/tex] that is parallel to [tex]$y = 2x+3$[/tex] and passes through [tex]$P(2,6)$[/tex] is:
[tex]$$
y = 2x + 2.
$$[/tex]
2. The slope of the line [tex]$L_2$[/tex], which is perpendicular to [tex]$L_1$[/tex], is:
[tex]$$
-\frac{1}{2},
$$[/tex]
and its equation, using point [tex]$P(2,6)$[/tex], is:
[tex]$$
y = -\frac{1}{2}x + 7.
$$[/tex]
