A line through the point (2,5) forms a right triangle with the x-axis and y-axis in the 1st quadrant. Determine the equation of the line that
will minimize the area of this triangle.

9514 1404 393

  5x +2y = 20

Step-by-step explanation:

Suppose 'a' is the x-intercept of the line.

Then the slope is given by the slope formula ...

  m = (y2 -y1)/(x2 -x1)

  m = (0 -5)/(a -2)

And the y-intercept is ...

  b = y -mx = 5 -(-5/(a-2))(2) = 5 +10/(a -2)

The area of the triangle is half the product of 'a' and 'b', so is ...

  A = 1/2ab

  A = 1/2(a)(5 +10/(a -2)) = 2.5a +5a/(a-2)

The area is minimized when the derivative of this is zero.

  A' = 2.5 -10/(a -2)^2 = 0

Solving for 'a' gives ...

  (a -2)^2 = 10/2.5 = 4

  a = 2 + 2 = 4 . . . . . . . . . . . x-intercept

  b = 5 +10/(4 -2) = 10 . . . . . y-intercept

  m = -5/(4 -2) = -5/2 . . . . . . slope

An equation of the line is ...

  y = -5/2x +10 . . . . . slope-intercept form

  5x +2y = 20 . . . . . standard form

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You may have noticed that the x-and y-intercepts are double the x- and y-values of the given point. That is, the given point is the midpoint of the hypotenuse of the triangle with minimum area. That is the general solution of this sort of problem.