Ms Taylor’s students give cards to each of the 3 class parent helpers there are 24 cards how many cards will each helper get if
1 answer:
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Answer:
Step-by-step explanation:
Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )
The average of x-coordinate will be:
<u>1) Finding </u>
- Average of the x coordinates:
- Average of the y coordinates:
similarly for y
<u>2) Finding the line through with slope m.</u>
Given a point and a slope, the equation of a line can be found using:
in our case this will be
this is our equation of the line!
<u>3) Find the squared vertical distances between this line and the three points.</u>
So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.
- Distance from point (1,3)
We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?
we'll go back to our equation of the line and use x=1.
now we know the two points at x=1: (1,3) and (1,-m+4)
to find the vertical distance we'll subtract the y-coordinates of each point.
finally, as asked, we'll square the distance
- Distance from point (2,3)
we'll do the same as above here:
vertical distance between the two points: (2,3) and (2,4)
squaring:
- Distance from point (3,6)
vertical distance between the two points: (3,6) and (3,m+4)
squaring:
3) Add up all the squared distances, we'll call this value R.
<u>4) Find the value of m that makes R minimum.</u>
Looking at the equation above, we can tell that R is a function of m:
you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)
now to find the minimum value we'll just use a condition that
now solve for m:
This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!
