Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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].
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].
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
