Certainly! To determine the value of [tex]\(k\)[/tex] so that the two lines given by the equations [tex]\(5x - 3y = -2\)[/tex] and [tex]\(x - 3ky + 7 = 0\)[/tex] are perpendicular, we can follow these steps:
1. Convert the equations to slope-intercept form (y = mx + b):
- For the first line [tex]\(5x - 3y = -2\)[/tex], we solve for [tex]\(y\)[/tex]:
[tex]\[
5x - 3y = -2 \implies -3y = -5x - 2 \implies y = \frac{5}{3}x + \frac{2}{3}
\][/tex]
Therefore, the slope of the first line, [tex]\(m_1\)[/tex], is [tex]\(\frac{5}{3}\)[/tex].
- For the second line [tex]\(x - 3ky + 7 = 0\)[/tex], we solve for [tex]\(y\)[/tex]:
[tex]\[
x - 3ky + 7 = 0 \implies -3ky = -x - 7 \implies y = \frac{1}{3k}x + \frac{7}{3k}
\][/tex]
Therefore, the slope of the second line, [tex]\(m_2\)[/tex], is [tex]\(\frac{1}{3k}\)[/tex].
2. Use the condition for perpendicular lines:
- Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]:
[tex]\[
m_1 \cdot m_2 = -1
\][/tex]
Substituting the slopes from above, we get:
[tex]\[
\left(\frac{5}{3}\right) \cdot \left(\frac{1}{3k}\right) = -1
\][/tex]
Simplify this equation:
[tex]\[
\frac{5}{3} \cdot \frac{1}{3k} = -1 \implies \frac{5}{9k} = -1
\][/tex]
3. Solve for [tex]\(k\)[/tex]:
- To find [tex]\(k\)[/tex], we solve the equation:
[tex]\[
\frac{5}{9k} = -1 \implies 5 = -9k \implies k = -\frac{5}{9}
\][/tex]
Therefore, the value of [tex]\(k\)[/tex] that makes the two lines perpendicular is:
[tex]\[
\boxed{-\frac{5}{9}}
\][/tex]
