To find the value of [tex]\( k \)[/tex] such that the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are perpendicular, we need to follow these steps:
1. Identify the slope of [tex]\( L_1 \)[/tex]:
The equation of [tex]\( L_1 \)[/tex] is given in slope-intercept form:
[tex]\[
y = 2x - 5
\][/tex]
In this form, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. Thus, the slope of [tex]\( L_1 \)[/tex] is:
[tex]\[
m_1 = 2
\][/tex]
2. Determine the condition for perpendicularity:
For two lines to be perpendicular, the product of their slopes must equal [tex]\(-1\)[/tex]. If the slope of [tex]\( L_2 \)[/tex] is [tex]\( m_2 \)[/tex], then:
[tex]\[
m_1 \cdot m_2 = -1
\][/tex]
Given [tex]\( m_1 = 2 \)[/tex], the slope of [tex]\( L_2 \)[/tex], [tex]\( m_2 \)[/tex], must satisfy:
[tex]\[
2 \cdot m_2 = -1 \implies m_2 = -\frac{1}{2}
\][/tex]
3. Find the slope of [tex]\( L_2 \)[/tex]:
Next, we need to rewrite the equation of [tex]\( L_2 \)[/tex] in slope-intercept form to identify its slope. The given equation is:
[tex]\[
6y + kx - 12 = 0
\][/tex]
Let's solve this equation for [tex]\( y \)[/tex] to get it in the form [tex]\( y = mx + b \)[/tex]:
[tex]\[
6y = -kx + 12
\][/tex]
[tex]\[
y = -\frac{k}{6}x + 2
\][/tex]
Thus, the slope [tex]\( m_2 \)[/tex] of [tex]\( L_2 \)[/tex] is:
[tex]\[
m_2 = -\frac{k}{6}
\][/tex]
4. Set up the equation for [tex]\( k \)[/tex]:
We determined that the slope of [tex]\( L_2 \)[/tex] must be [tex]\(-\frac{1}{2}\)[/tex]. Therefore, we set:
[tex]\[
-\frac{k}{6} = -\frac{1}{2}
\][/tex]
5. Solve for [tex]\( k \)[/tex]:
To solve for [tex]\( k \)[/tex], first eliminate the negative sign by multiplying both sides by [tex]\(-1\)[/tex]:
[tex]\[
\frac{k}{6} = \frac{1}{2}
\][/tex]
Next, solve for [tex]\( k \)[/tex] by multiplying both sides by 6:
[tex]\[
k = 6 \cdot \frac{1}{2}
\][/tex]
[tex]\[
k = 3
\][/tex]
Thus, the value of [tex]\( k \)[/tex] such that the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are perpendicular is [tex]\( \boxed{3} \)[/tex].
