Answer :
Answer:
The line that passes through (-5,2) and is perpendicular to y+3=2x is [tex]y=\frac{-x}{2}+\frac{-1}{2}[/tex]
Solution:
Given, line equation is y + 3 = 2x ⇒ 2x – y – 3 = 0
We have to find a line that is perpendicular to y + 3 = 2x and passing through (-5, 2).
Now, let us find the slope of the given line,
[tex]\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{-1}=2[/tex]
We know that, slope of a line [tex]\times[/tex] slope of perpendicular line = -1
Then, 2 [tex]\times[/tex] slope of perpendicular line = -1
[tex]\rightarrow \text { slope of perpendicular line }=-1 \times \frac{1}{2}=-\frac{1}{2}[/tex]
Now, slope of our required line = [tex]\frac{-1}{2}[/tex] and it passes through (-5, 2)
We know that, point slope form is
[tex]y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }[/tex]
Here in our problem, [tex]m=-\frac{1}{2}, \text { and }\left(x_{1}, y_{1}\right)=(-5,2)[/tex]
Then, line equation [tex]\rightarrow y-2=-\frac{1}{2}(x-(-5))[/tex]
2(y – 2) = -1(x + 5)
2y – 4 = -x – 5
x + 2y + 5 – 4 = 0
[tex]y=\frac{-x}{2}+\frac{-1}{2}[/tex]
Thus the point slope form is [tex]y=\frac{-x}{2}+\frac{-1}{2}[/tex]