Respuesta :
Answer:
The equation of a line perpendicular to the given line and passes through the given point is:
[tex]y=-3x+15[/tex]Explanation:
Given the line:
[tex]y+2=\frac{1}{3}(x-5)_{}[/tex]This can be rewritten as:
[tex]\begin{gathered} y=\frac{1}{3}x-5-2 \\ \\ y=\frac{1}{3}x-7 \end{gathered}[/tex]This is a line with slope 1/3, and y-intercept -7.
A line perpendicular to this line has it's slope as the negative reciprocal of 1/3.
The negativen reciprocal of 1/3 is -3.
So, the line is of the form:
[tex]y=-3x+b[/tex]We can find the y-intercept b, by using the given point, (4, 3) in the perpendicular equation we formend.
Where x = 4, and y = 3
Doing this, we have:
[tex]\begin{gathered} 3=-3(4)+b \\ 3=-12+b \end{gathered}[/tex]Add 12 to both sides
[tex]b=3+12=15[/tex]Therefore, the equation of a line perpendicular to the given line and passes through the given point is:
[tex]y=-3x+15[/tex]