Respuesta :
If two lines are perpendicular, then their slopes have a product of -1 or you could say they are opposite signed reciprocals of each other.
The given equation has a slope of -3. We know this because it is in slope-intercept form y = mx + b ...y = (slope)x + (y-intercept).
so taking the slope -3...the opposite signed reciprocal would be +[tex] \frac{1}{3} [/tex]
Now we have a slope and a point... we can use them to find the b (y-intercept)
x = 4, y = 3, and m = [tex] \frac{1}{3} [/tex]
[tex]3= \frac{1}{3} (4)+b[/tex]
[tex]3= \frac{4}{3} +b[/tex]
Using the common denominator fraction for 3, we subtract 4/3 from both sides
[tex]3= \frac{9}{3} [/tex]
[tex] \frac{5}{3} =b[/tex]
plug the values for m and b into the slope intercept form
[tex]y= \frac{1}{3} x+ \frac{5}{3} [/tex]
The given equation has a slope of -3. We know this because it is in slope-intercept form y = mx + b ...y = (slope)x + (y-intercept).
so taking the slope -3...the opposite signed reciprocal would be +[tex] \frac{1}{3} [/tex]
Now we have a slope and a point... we can use them to find the b (y-intercept)
x = 4, y = 3, and m = [tex] \frac{1}{3} [/tex]
[tex]3= \frac{1}{3} (4)+b[/tex]
[tex]3= \frac{4}{3} +b[/tex]
Using the common denominator fraction for 3, we subtract 4/3 from both sides
[tex]3= \frac{9}{3} [/tex]
[tex] \frac{5}{3} =b[/tex]
plug the values for m and b into the slope intercept form
[tex]y= \frac{1}{3} x+ \frac{5}{3} [/tex]
