How do you write the equation of a line in slope intercept, point slope and standard form given P: (3, -4) and Q: (2,1)?

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Answer 1 · 2017-08-05T04:50:45

Slope intercept form: #y=-5x+11#
Point slope form: #(y+4)=-5(x-3)#
Standard form: #-5x+y=11#

First step is to find the slope (gradient), #m#:

#m=(rise)/(run)=(y_2-y_1)/(x_2-x_1) =(1-(-4))/(2-3)=5/(-1)=-5#

To write the line in 'point slope' form we can choose either point and substitute it into this form:

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

We'll choose the first point, P:

#(y-(-4))=-5(x-3)#

Which simplifies to:

#(y+4)=-5(x-3)#

To write it in 'slope intercept' or 'standard' form, we need to find the y-intercept.

To do so we use one of the points in the form for slope intercept:

#y=mx+c#

Let's substitute in point Q to keep it interesting:

#1=-5(2)+c#

Rearranging, #c=11#.

We can immediately write the equation of the line in standard form:

#y=-5x+11#

Standard form is:

#ax+by=c#

We get to it by moving the #x# term in slope intercept form to the left of the equals sign:

#-5x+y=11#

And we're done!