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)?

Question

Charles Anctil · Accepted Answer

#4# 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#

Aurora Alvarado · Answer

undefined To find the minimum area of the triangle formed by the x-axis, y-axis, and a line passing through the point (2, 1), we need to consider the slope of the line.

The line intersects the x-axis at some point A and the y-axis at some point B. The coordinates of A will be (a, 0), and the coordinates of B will be (0, b). The line passes through the point (2, 1), so substituting these coordinates into the equation of the line will give us the slope.

Using the slope-intercept form of the equation of a line, which is $y = mx + b$, where m is the slope, we can find the slope of the line passing through the point (2, 1).

Then, we can find the equations of the lines AB and BC, where AB is parallel to the y-axis and BC is parallel to the x-axis. The coordinates of the intersection point of these lines will give us the coordinates of A and B.

After finding the coordinates of A and B, we can use the formula for the area of a triangle, which is $\frac{1}{2} \times base \times height$, to find the area of the triangle.

This minimum area will occur when the line is perpendicular to one of the axes. Thus, we find the perpendicular line to the given line that passes through (2, 1) and find the area of the triangle formed by this line with the axes.

Therefore, the minimum area of the triangle can be found using the above steps.