What is the minimum area of a triangle formed by the x-axis and the y-axis and a line through the point (2,1)?

1 Answer
Feb 10, 2018

#4#

Explanation:

For a line through #(2,1)#, let #b = #the #y#-intercept and #a = #the #x#-intercept.

The area of the triangle is #A = 1/2ab#

To get one variable, we need to find #a# in terms of #b# or vice versa.

We know that the #y#-intercept is #(0,b)# and we know that the point #(2,1)# is on the line, so either use

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

or use the the fact that the equation #y=mx+b# is to be true at #(2,1)# to get:

#m = (1-b)/2#

So the line has equation: #y = (1-b)/2 x +b#.

The #x# intercept, which we have called #a# solves #0 = (1-b)/2 a +b#.

So, #a = (2b)/(b-1)#

Area #A = 1/2(b)((2b)/(b-1)) = b^2/(b-1)#

From here, find and test the critical numbers to see that #A# is minimum when #b=2#.

And calculate that, when #b=2#, #A = 4#