Minimize distance between three points.

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To minimize the sum of the squares of the vertical distances from the points (1,1), (2,2), and (3,2) to the line y=mx, the focus should be on the y-values since the x-values will cancel out. The vertical distance from a point (x1, y1) to the line can be expressed as the difference between y1 and mx, which simplifies the problem to a function of m. By summing the squared vertical distances for the three points, an equation can be formed that depends solely on m. The goal is to find the value of m that minimizes this sum. This approach leverages single variable calculus to find the optimal slope for the line.
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Find the value of m such that the sum of the squares of the vertical distances from each of the points (1,1) , (2,2) , and (3,2) to the line y=mx is minimized. Hint: Find the sum as a function of m (no x in the expression) and then minimize it.



Distance equation. d=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}



So, our professor made it easier on use and since we have single variable calculus the distance to be found is vertical to the line y=mx. Therefore, the x values will cancel out and we will only be interested in the y values. With this knowledge I set out and tried to set up an equation for the distance formula for each of the values given; but, x values were given in the equation I used for the distance values when I subbed the y value for mx. I'm not sure how to juggle these three equations. How would you proceed with solving this equation. Thank you.
 
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Can you show what you did?

I'm not sure how to juggle these three equations.
Add them. The sum of the squared distances should be minimized.
 
How would you express the vertical distance between a given point (x_{1},y_{i}) and a point on the line y=mx. (hint, what is the vertical distance between the y values ?).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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