Option 2
The equation of a line that is perpendicular to the given line [tex]y - 3 = \frac{8}{3}(x + 2)[/tex] and that passes through the given point (-2, 3) is [tex]y-3=\frac{-3}{8}(x+2)[/tex]
Given that line that is perpendicular to the given line [tex]y - 3 = \frac{8}{3}(x + 2)[/tex] and that passes through the given point (-2, 3)
We have to find the equation of line
The point slope form is given as:
[tex]y - y_1 = m(x - x_1)[/tex]
Where "m" is the slope of line
Comparing the given equation [tex]y - 3 = \frac{8}{3}(x + 2)[/tex] with point slope form [tex]y - y_1 = m(x - x_1)[/tex]
[tex]m = \frac{8}{3}[/tex]
Thus slope of line is [tex]m = \frac{8}{3}[/tex]
We know that product of slopes of perpendicular lines are always equal to -1
Slope of given line x slope of line perpendicular to given line = -1
[tex]\frac{8}{3} \times \text { slope of line perpendicular to given line }=-1[/tex]
[tex]\begin{array}{l}{\text { slope of line perpendicular to given line }=-1 \times \frac{3}{8}} \\\\ {\text { slope of line perpendicular to given line }=\frac{-3}{8}}\end{array}[/tex]
Now we have to find the equation of line having slope [tex]m = \frac{-3}{8}[/tex] and passes through point (-2, 3)
[tex]\text {substitute } m=\frac{-3}{8} \text { and }\left(x_{1}, y_{1}\right)=(-2,3) \text { in point slope form }[/tex]
[tex]y - y_1 = m(x - x_1)[/tex]
[tex]\begin{array}{l}{y-(3)=\frac{-3}{8}(x-(-2))} \\\\ {y-3=\frac{-3}{8}(x+2)}\end{array}[/tex]
Thus the required equation is found and option 2 is correct
