Question: For what value of k, the line kx-3y+6=0 is perpendicular to the line joining (4,3) and (5,-3)?

Solution:

Here, equation of line 1 is (kx - 3y +6 = 0)
Slope of line 1 ($m_1$) = $\dfrac{- coefficient\;of\;x}{coefficient\;of\;y}$
$= \dfrac{- k}{-3}$
$= \dfrac{k}{3}$

For line 2, two points are (4,3) and (5,-3).
Let (4,3) be ($x_1,y_1$) and (5,-3) be ($x_2,y_2$).

Slope of line 2 ($m_2$) = $\dfrac{y_2-y_1}{x_2 - x_1}$
$= \dfrac{-3 -3}{5-4}$
$= \dfrac{-6}{1}$
$= -9$

Given, two lines are perpendicular to each other,
So, $m_1×m_2 = -1$

$or, \dfrac{k}{3} × (-6) = -1$

$or, \dfrac{-6k}{3} = -1$

$or, \dfrac{-3k}{1} = -1$

$or, -2k = -1$

$or, 2k = 1$

$\therefore k = \dfrac{1}{2}$

Hence, the required value of k is $\frac{1}{2}$ for the line kx-3y+6=0 to be perpendicular with the other given line.