Answer:
First option.
Third option.
Fifth option.
Step-by-step explanation:
The equation of the line in Slope-Intercept form is:
[tex]y=mx+b[/tex]
Where "m" is the slope and "b" is the y-intercept.
The equation of the line in Point-slope form is:
[tex]y - y_1 = m(x-x_1)[/tex]
Where "m" is the slope and [tex](x_1,y_1)[/tex] is a point on the line.
By definition, the slopes of perpendicular line are negative reciprocals.
Then, given the line:
[tex]y- 1 = \frac{1}{3}(x+2)[/tex]
We know that a line perpendicular to it, must have this slope:
[tex]m=-3[/tex]
Let's check each option:
1) [tex]y + 2 = -3(x -4)[/tex]
Since [tex]m=-3[/tex], this line is perpendicular to the line [tex]y- 1 = \frac{1}{3}(x+2)[/tex]
2) [tex]y - 5 = 3(x + 11)[/tex]
Since [tex]m\neq -3[/tex], this line is not perpendicular to the line [tex]y- 1 = \frac{1}{3}(x+2)[/tex]
3) [tex]y = -3x-\frac{5}{3}[/tex]
Since [tex]m=-3[/tex], this line is perpendicular to the line [tex]y- 1 = \frac{1}{3}(x+2)[/tex]
4) [tex]\frac{1}{3}x - 2[/tex]
Since [tex]m\neq -3[/tex], this line is not perpendicular to the line [tex]y- 1 = \frac{1}{3}(x+2)[/tex]
5) [tex]3x + y = 7[/tex]
Solving for "y":
[tex]y =-3x+ 7[/tex]
Since [tex]m=-3[/tex], this line is perpendicular to the line [tex]y- 1 = \frac{1}{3}(x+2)[/tex]
