Solution:
The equation of a line that passes through a point is expressed as
[tex]\begin{gathered} y-y_1=m(x-x_1)\text{ ---- equation 1} \\ where \\ m\Rightarrow slope\text{ of the line} \\ y\Rightarrow y-intercept\text{ of the line} \\ (x_1,y_1)\Rightarrow coordinate\text{ of the point through which the line passes} \end{gathered}[/tex]Two lines A and B are said to be perpendicular if the slope of line A is equal to the negative reciprocal of the slope of line B.
Thus, lines A and B are perpendicular if
[tex]m_A=-\frac{1}{m_B}\text{ ---- equation 2}[/tex]Let the line equation
[tex]y-4=\frac{2}{5}(x-6)[/tex]represent the line A.
step 1: Evaluate the slope of line A.
Comparing the equation of line A with equation 1, we can conclude that the slope of the line A is
[tex]m_A=\frac{2}{5}[/tex]step 2: Evaluate the slope of line B.
Since lines A and B are perpendicular, we have
[tex]\begin{gathered} From\text{ equation 2,} \\ \begin{equation*} m_A=-\frac{1}{m_B} \end{equation*} \\ where \\ m_A=\frac{2}{5} \\ thus, \\ \frac{2}{5}=-\frac{1}{m_B} \\ cross-multiply \\ -2m_B=5 \\ divide\text{ both sides by -2} \\ \frac{-2m_B}{-2}=\frac{5}{-2} \\ \Rightarrow m_B=-\frac{5}{2} \end{gathered}[/tex]step 3: Evaluate the line B.
Since the line B passes through the points (-3,2), recall from equation 1
[tex]\begin{equation*} y-y_1=m(x-x_1)\text{ } \end{equation*}[/tex]where
[tex]\begin{gathered} x_1=-3 \\ y_1=2 \\ m=m_B=-\frac{5}{2} \end{gathered}[/tex]Substitute these values into equation 1.
Thus,
[tex]\begin{gathered} y-2=-\frac{5}{2}(x-(-3)) \\ \Rightarrow y-2=-\frac{5}{2}(x+3) \end{gathered}[/tex]Hence, the equation of the line is
[tex]y-2=-\frac{5}{2}(x+3)[/tex]The correct option is A
