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Sagot :
To determine the value of [tex]\( k \)[/tex] for which the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are perpendicular, we need to follow these steps:
1. Identify the slope of Line [tex]\( L_1 \)[/tex]:
The equation of line [tex]\( L_1 \)[/tex] is given as [tex]\( y = 2x - 5 \)[/tex]. This is in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
Thus, the slope of line [tex]\( L_1 \)[/tex] is:
[tex]\[ m_1 = 2 \][/tex]
2. Determine the slope condition for perpendicularity:
Recall that if two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]. Let [tex]\( m_2 \)[/tex] be the slope of [tex]\( L_2 \)[/tex]. Thus:
[tex]\[ m_1 \cdot m_2 = -1 \][/tex]
Substituting [tex]\( m_1 = 2 \)[/tex], we get:
[tex]\[ 2 \cdot m_2 = -1 \][/tex]
Solving for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = -\frac{1}{2} \][/tex]
3. Convert the equation of Line [tex]\( L_2 \)[/tex] to slope-intercept form:
The given equation of line [tex]\( L_2 \)[/tex] is [tex]\( 6y + kx - 12 = 0 \)[/tex]. To find its slope, we need to rewrite this equation in the form [tex]\( y = mx + b \)[/tex]:
[tex]\[ 6y + kx - 12 = 0 \][/tex]
Rewrite it to express [tex]\( y \)[/tex] on one side:
[tex]\[ 6y = -kx + 12 \][/tex]
Divide every term by 6 to isolate [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{k}{6}x + 2 \][/tex]
Here, the slope [tex]\( m_2 \)[/tex] of line [tex]\( L_2 \)[/tex] is [tex]\( -\frac{k}{6} \)[/tex].
4. Set up the equation for perpendicular slopes:
For the lines to be perpendicular, we previously determined [tex]\( m_2 = -\frac{1}{2} \)[/tex]. So,
[tex]\[ -\frac{k}{6} = -\frac{1}{2} \][/tex]
5. Solve for [tex]\( k \)[/tex]:
Equate the slopes:
[tex]\[ -\frac{k}{6} = -\frac{1}{2} \][/tex]
Remove the negative signs and multiply both sides by 6:
[tex]\[ k = \frac{1}{2} \times 6 \][/tex]
Simplify the right side:
[tex]\[ k = 3 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] for which the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are perpendicular is:
[tex]\[ \boxed{3} \][/tex]
1. Identify the slope of Line [tex]\( L_1 \)[/tex]:
The equation of line [tex]\( L_1 \)[/tex] is given as [tex]\( y = 2x - 5 \)[/tex]. This is in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
Thus, the slope of line [tex]\( L_1 \)[/tex] is:
[tex]\[ m_1 = 2 \][/tex]
2. Determine the slope condition for perpendicularity:
Recall that if two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]. Let [tex]\( m_2 \)[/tex] be the slope of [tex]\( L_2 \)[/tex]. Thus:
[tex]\[ m_1 \cdot m_2 = -1 \][/tex]
Substituting [tex]\( m_1 = 2 \)[/tex], we get:
[tex]\[ 2 \cdot m_2 = -1 \][/tex]
Solving for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = -\frac{1}{2} \][/tex]
3. Convert the equation of Line [tex]\( L_2 \)[/tex] to slope-intercept form:
The given equation of line [tex]\( L_2 \)[/tex] is [tex]\( 6y + kx - 12 = 0 \)[/tex]. To find its slope, we need to rewrite this equation in the form [tex]\( y = mx + b \)[/tex]:
[tex]\[ 6y + kx - 12 = 0 \][/tex]
Rewrite it to express [tex]\( y \)[/tex] on one side:
[tex]\[ 6y = -kx + 12 \][/tex]
Divide every term by 6 to isolate [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{k}{6}x + 2 \][/tex]
Here, the slope [tex]\( m_2 \)[/tex] of line [tex]\( L_2 \)[/tex] is [tex]\( -\frac{k}{6} \)[/tex].
4. Set up the equation for perpendicular slopes:
For the lines to be perpendicular, we previously determined [tex]\( m_2 = -\frac{1}{2} \)[/tex]. So,
[tex]\[ -\frac{k}{6} = -\frac{1}{2} \][/tex]
5. Solve for [tex]\( k \)[/tex]:
Equate the slopes:
[tex]\[ -\frac{k}{6} = -\frac{1}{2} \][/tex]
Remove the negative signs and multiply both sides by 6:
[tex]\[ k = \frac{1}{2} \times 6 \][/tex]
Simplify the right side:
[tex]\[ k = 3 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] for which the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are perpendicular is:
[tex]\[ \boxed{3} \][/tex]
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