What is the equation of the line with slope # m= 19/25 # that passes through # (16/5 73/10) #?

Question

What is the equation of the line with slope # m= 19/25 # that passes through # (16/5  73/10) #?

Abigail Allen · Accepted Answer

#y-73/10=19/25(x-16/5)larr# Point-slope form

#y=19/25x+1217/250larr# y=mx+b form

#-19/25x+y=1217/250larr# Standard form

#x-304/125#-73/10=19/25# #ycancel(-73/10+73/10) = 19/25x - 304/125+73/10 # [304/125(2/2)]+[73/10(25/25)]#y=19/25x- #y=1825/250+608/250x-19/25x# In the form of y=mx+b, the equation is #y=19/25x+1217/250larr#. The equation can also be written in standard form, which is #Ax+By=C#. #-19/25x + y = cancel(19/25x -19/25x) + 1217/250# #-19/25x+y=1217/250larr# is the standard form of the equation.

Ethan Adkins · Answer

undefined The equation of the line with slope $ m = \frac{19}{25} $ that passes through the point $ \left(\frac{16}{5}, \frac{73}{10}\right) $ is:

$$ y - \frac{73}{10} = \frac{19}{25}(x - \frac{16}{5}) $$

Genesis Allen · Answer

undefined The equation of the line with slope $m = \frac{19}{25}$ that passes through the point $\left(\frac{16}{5}, \frac{73}{10}\right)$ is:

$$y - y_1 = m(x - x_1)$$

Substitute $m = \frac{19}{25}$, $x_1 = \frac{16}{5}$, and $y_1 = \frac{73}{10}$ into the equation:

$$y - \frac{73}{10} = \frac{19}{25}\left(x - \frac{16}{5}\right)$$

Simplify the equation:

$$y - \frac{73}{10} = \frac{19}{25}x - \frac{19}{5}$$

$$y = \frac{19}{25}x - \frac{19}{5} + \frac{73}{10}$$

$$y = \frac{19}{25}x - \frac{19}{5} + \frac{146}{20}$$

$$y = \frac{19}{25}x - \frac{19}{5} + \frac{73}{10}$$

$$y = \frac{19}{25}x - \frac{38}{10} + \frac{73}{10}$$

$$y = \frac{19}{25}x + \frac{35}{10}$$

$$y = \frac{19}{25}x + \frac{7}{2}$$

The equation of the line is $y = \frac{19}{25}x + \frac{7}{2}$.