There are different methods to approach this:

#1.#. Form simultaneous equations based on #y =mx + c#

(Substitute the values of #x and y# which have been given.)

# -34 = m(3) + c# and #-9 = m(4) + c#

Solve them to find the values of #m and c#, which will give the equation of the line. Elimination by subtracting the 2 equations is probably the easiest as the #c# terms will subtract to 0.

#2.# Use the two points to find the gradient. #m = (y_2 - y_1)/(x_2 - x_1)#

Then substitute values for #m# and one point #x, y# into #y =mx + c# to find #c#.

Finally answer in the form #y =mx + c#, using the values for #m and c# you have found.

#3.# Use the formula from coordinate (or analytical) geometry which uses 2 points and a general point #(x, y)#

#(y - y_1)/(x - x_1) = (y_2 - y_1)/(x_2 - x_1)#

Substitute the values for the 2 given points, calculate the fraction on the right hand side (which gives the gradient), cross-multiply and with a small amount of transposing, the equation of the line is obtained.

#(y - (-34))/(x - 3) = (-9 - (-34))/(4 - 3) = 25/1#

#(y+34)/(x-3) = 25/1# Now cross-multiply

#y+34 = 25x-75#

# y = 25x -109#