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

1 Answer
Anthony R.
Aug 17, 2017

#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

Explanation:

Seeing how we already have the slope and a coordinate, we can find the equation of the line by using the point-slope formula: #y-y_1=m(x-x_1)# where #m# is the slope #(m=19/25)# and #(x_1, y_1)# is a point on the line. Thus, #(16/5,73/10)-> (x_1,y_1)#.

The equation is then...

#y-73/10=19/25(x-16/5)#

...in point slope form.

Since you did not specify in what form the equation should be expressed, the above is an acceptable answer but we could also rewrite the equation is #y=mx+b# form. To do this, we solve for #y#.

#y-73/10=19/25x-304/125#

#ycancel(-73/10+73/10)=19/25x-304/125+73/10#

#y=19/25x-[304/125(2/2)]+[73/10(25/25)]#

#y=19/25x-608/250+1825/250#

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

Alternatively, the equation could also be expressed in standard form: #Ax+By=C#

#-19/25x+y=cancel(19/25x-19/25x)+1217/250#

#-19/25x+y=1217/250larr# The equation is standard form

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