How do you write the point-slope form of the equation below that represents the line that passes through the points (−3, 2) and (2, 1)?
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#slope=(y_2-y_1)/(x_2-x_1)#
#y=-1/5x+7/5#
The slope of the line passing through the two points #(-3,2)# and #(2,1)# is given by
#slope=(y_2-y_1)/(x_2-x_1)#
#slope=(1-2)/(2-(-3))#
#slope =-1/5#
The equation of the straight line is
#y=a*x+b#
#slope = a=-1/5#
#y=-1/5x+b#
The point #(2,1)# is a point on the straight line. Plugging the coordinates of this point and the line equation allows us to find the Y-intercept.
#1=-1/5xx2+b#
#b=1+2/5#
#b=7/5#
The equation of the straight-line will be
#y=-1/5x+7/5#
To find the X-intercept assign the value 0 to #y#
#0=-1/5x+7/5#
#1/5x=7/5#
#x=7#
graph{-1/5*x+7/5 [-9.63, 10.37, -3.4, 6.6]}
There are several ways of answering this question:
Method 1 . Use the two points to find the gradient #m#.
#m = (y_2 - y_1)/(x_2 - x_1)" "# Then use one of the points as #(x,y)#
Substitute #m,x and y# into #y = mx + c# and solve to find #c#.
Now use the values for #m and c# in #y = mx + c# to find the required equation.
This method is fine, but it requires several substitutions and it is easy to lose track of where you are.
Method 2 . Use the two points as #x and y# and substitute each into #y = mx + c " " # This forms two equations which can be solved simultaneously to find m and c.
Method 3 . Use the two points as #(x_1,y_1) and (x_2, y_2)# and substitute into the formula for finding the equation of a line if 2 points are known: # " "(y - y_1)/(x - x_1) = (y_2 - y_1)/(x_2 - x_1)#
# " "(y - 1)/(x - 2) = (2 - 1)/(-3 - 2)#
# " "(y - 1)/(x - 2) = 1/-5 " now cross-multiply"#
#-5(y - 1) = (x - 2) " solve for y"#
#-5y + 5 = x - 2#
#-5y " "= x -2 -5#
#y = (-1x)/5 + 7/5 " divide by -5"#
