What is the equation in point-slope form of the line passing through (–2, 1) and (4, 13)?

The Point-Slope form of the Equation of a Straight Line is:

# (y-k)=m*(x-h) #
#m# is the Slope of the Line
#(h,k)# are the co-ordinates of any point on that Line.

• To find the Equation of the Line in Point-Slope form, we first need to Determine it's Slope . Finding the Slope is easy if we are given the coordinates of two points.

Slope(#m#) = #(y_2-y_1)/(x_2-x_1)# where #(x_1,y_1)# and #(x_2,y_2)# are the coordinates of any two points on the Line

The coordinates given are #(-2,1)# and #(4,13)#

Slope(#m#) = #(13-1)/(4-(-2))# = #12/6# = #2#

• Once the Slope is determined, pick any point on that line. Say #(-2,1)#, and Substitute it's co-ordinates in #(h,k)# of the Point-Slope Form.

We get the Point-Slope form of the equation of this line as:

#(y-1)=(2)*(x-(-2))#

• Once we arrive at the Point-Slope form of the Equation, it would be a good idea to Verify our answer. We take the other point #(4,13)#, and substitute it in our answer.

#(y-1) = 13-1 = 12#

#(2)*(x-(-2)) = (2)*(4-(-2)) = 2*6 = 12#

As the left hand side of the equation is equal to the right hand side, we can be sure that the point #(4,13)# does lie on the line.

• The graph of the line would look like this:
  graph{2x-y=-5 [-10, 10, -5, 5]}