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mr_godi [17]
2 years ago
10

Need helps pls will give thanks

Mathematics
1 answer:
inessss [21]2 years ago
4 0
B- P’s gradient is greater than Q’s
Hope this helps!
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What is the measure of angle A?
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actual answer is 77.32 !

Step-by-step explanation:


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Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y
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Answer:

m=\dfrac{3}{2}

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}

<u>1) Finding (\overline{x},\overline{y})</u>

  • Average of the x coordinates:

\overline{x} = \dfrac{1+2+3}{3}

\overline{x} = 2

  • Average of the y coordinates:

similarly for y

\overline{y} = \dfrac{3+3+6}{3}

\overline{y} = 4

<u>2) Finding the line through (\overline{x},\overline{y}) with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

(y-y_1)=m(x-x_1)

in our case this will be

(y-\overline{y})=m(x-\overline{x})

(y-4)=m(x-2)

y=mx-2m+4

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.

y=m(1)-2m+4

y=-m+4

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.

d_1=3-(-m+4)

d_1=m-1

finally, as asked, we'll square the distance

(d_1)^2=(m-1)^2

  • Distance from point (2,3)

we'll do the same as above here:

y=m(2)-2m+4

y=4

vertical distance between the two points: (2,3) and (2,4)

d_2=3-4

d_2=-1

squaring:

(d_2)^2=1

  • Distance from point (3,6)

y=m(3)-2m+4

y=m+4

vertical distance between the two points: (3,6) and (3,m+4)

d_3=6-(m+4)

d_3=2-m

squaring:

(d_3)^2=(2-m)^2

3) Add up all the squared distances, we'll call this value R.

R=(d_1)^2+(d_2)^2+(d_3)^2

R=(m-1)^2+4+(2-m)^2

<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:

R(m)=(m-1)^2+4+(2-m)^2

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)

\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)

\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)

now to find the minimum value we'll just use a condition that \dfrac{dR}{dm}=0

0=2(m-1)+2(2-m)(-1)

now solve for m:

0=2m-2-4+2m

m=\dfrac{3}{2}

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!

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