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