Answer:
[tex]y=\displaystyle\frac{1}{2} x+\displaystyle \frac{5}{2}[/tex]
Step-by-step explanation:
Hi there!
Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept.
Perpendicular lines always have slopes that are negative reciprocals (ex. 1/2 and -2, 3/4 and -4/3)
Determine the slope (m):
[tex]y = 1 -2x[/tex]
Rearrange into slope-intercept form:
[tex]y = -2x+1[/tex]
Now, we can identify clearly that the slope is -2. Because perpendicular lines always have slopes that are negative reciprocals, a perpendicular line would have a slope of [tex]\displaystyle\frac{1}{2}[/tex]. Plug this into [tex]y=mx+b[/tex]:
[tex]y=\displaystyle\frac{1}{2} x+b[/tex]
Determine the y-intercept (b):
[tex]y=\displaystyle\frac{1}{2} x+b[/tex]
Plug in the given point (1,3) and solve for b:
[tex]3=\displaystyle\frac{1}{2} *1+b\\\\b=\displaystyle \frac{5}{2}[/tex]
Therefore, the y-intercept is [tex]\displaystyle \frac{5}{2}[/tex]. Plug this back into [tex]y=\displaystyle\frac{1}{2} x+b[/tex]:
[tex]y=\displaystyle\frac{1}{2} x+\displaystyle \frac{5}{2}[/tex]
I hope this helps!
