Answer-
[tex]\mathbf{x + 2y = 0}[/tex] has a slope of [tex]-\dfrac{1}{2}[/tex]
Solution-
The general point slope formula of straight line,
[tex]y=mx+c[/tex]
Where,
m = slope
c = y-intercept
[tex]\mathbf{x + 2y = 0}[/tex]
[tex]\Rightarrow y=-\dfrac{x}{2}[/tex]
Comparing with general equation, slope = [tex]-\dfrac{1}{2}[/tex]
[tex]\mathbf{x - 2y = 0}[/tex]
[tex]\Rightarrow y=\dfrac{x}{2}[/tex]
Comparing with general equation, slope = [tex]\dfrac{1}{2}[/tex]
[tex]\mathbf{-x + 2y = 0}[/tex]
[tex]\Rightarrow y=\dfrac{x}{2}[/tex]
Comparing with general equation, slope = [tex]\dfrac{1}{2}[/tex]
[tex]\mathbf{y = x -\dfrac{1}{2}}[/tex]
Comparing with general equation, slope = 1, y-intercept = [tex]-\dfrac{1}{2}[/tex]
Therefore, [tex]\mathbf{x + 2y = 0}[/tex] has a slope of [tex]-\dfrac{1}{2}[/tex]
