This is our last E/C for the semester: Find the equation of the shortest line co

Question

This is our last E/C for the semester: Find the equation of the shortest line connecting points (0, y), (2, 1), and (x, 0). First convince yourself that this problem has a solution - it does. Next, you may use alphawolfram.com to solve your equations. The command in this program is: solve [equation]-, for example, to solve x^3 - 6x^2 + 2x - 1 = 0, you type solve xA3 - 6*xA2 + 2*x - 1 = 0 and then click on the equal sign. Remember that x has to be a real number so you must discard and complex roots that you may obtain. In this example, the polynomial has one real and two complex roots. Its only real root the beautiful number x = 2 + 1/3^3 squareroot 351/2 - 3 squareroot 351/2 - 3 squareroot 1689/2 +^3 squareroot 1/2 (117 + squareroot 1689)/3^2/3 which is approximately equal to 5.679. You may toggle between exact and approximate values by clicking on the "approximate form" bar. Should your answer not be an integer, round ail numbers to two decimal places. Remember, you are looking for the equation of the line of shortest length so your final answer should look like y = mx + b. Your job is to find m and b so that the line has least length.

Explanation / Answer

if (2,1) is the midpoint of the points (0,y) and (x,0) then the line passing theough the given three points will be the shortest

=> (0+x)/2 = 2 , x = 4   and (y+0)/2 = 1, y = 2

=> the two intercepts become (4,0) and (0,2)

m = (2-0)/(0-4) = -1/2 and b = 2

=> the equation becomes y = -1/2*x + 2   or y = -x/2 + 2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.