How do you write an equation of the line that passes through (-3,4) and (1,0)?

2 Answers

#y = -x+1#

Explanation:

The line equation is of the form
#y = ax + b#

As the line passes through this two points
#(x_0,y_0) = (-3,4)#
#(x_1,y_1) = (1,0)#

they both obey the equation
#y_1 = ax_1 + b => 0 = a+b => a = -b#
#y_0 = ax_0 + b => 4 = -3a+b#

Therefore:
# 4 = -3a - a => 4 = -4a => a = -1 and b = 1#

So, the equation of the line that passes through those points is
#y = -x+1#

Sep 2, 2016

We will write the equation in slope-intercept form, #y=mx+c#. I have calculated below that the slope of the line is -1 and its y-intercept is 1.

The equation of the line can be written: #y=-x+1#

(we don't write the number '1' in front of the #x#, since #1x=x#)

Explanation:

First step: find the slope (gradient) of the line:

#m=(y_2-y_1)/(x_2-x_1)#

It doesn't matter which of the two given points we decided is 'Point 1' #(x_1,y_1)#, but let's choose the point #(1,0)#, so #x_1=1# and #y_1=0#.

Similarly for the other point, #(-3,4)#, so #x_2=-3# and #y_2=4#.

So #m=(y_2-y_1)/(x_2-x_1)=(4-0)/(-3-1) =4/(-4)=-1#

Now that we know the slope, we can use it and the value of one of the points we were given to find the y-intercept of the line. This is the point where the line crosses the y-axis. The y-axis is the line #x=0#, so if we substitute #m=-1# and, for example, #x_1=1# and #y_1=0#, into the equation:

#y=mx+c#

#0=-1(1)+c#

Rearranging, #c=1#.