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Answer:
The given points (-4,1) and (4,6) are used to find the equation of the line the cross through them. First, we have to find the slope:
[tex]m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }\\m=\frac{6-1}{4-(-4)}=\frac{5}{8}[/tex]
Now, we use point-slope formula to find the equation:
[tex]y-y_{1}=m(x-x_{1})\\ y-1=\frac{5}{8} (x-(-4))\\y=\frac{5}{8}x+\frac{20}{8}+1\\y=\frac{5}{8}x+\frac{20+8}{8}\\y=\frac{5}{8}x+\frac{28}{8} \\y=\frac{5}{8}x+\frac{7}{2}[/tex]
Now, we test each point to see which is on the line:
For (0, 3.5):
[tex]y=\frac{5}{8}x+\frac{7}{2}[/tex]
[tex]3.5=\frac{5}{8}(0)+\frac{7}{2}[/tex]
[tex]3.5=\frac{7}{2}[/tex]
[tex]3.5=3.5[/tex]
So, the first point is on the line, because it satisfies the liner equation.
For (12,11):
[tex]11=\frac{5}{8}(12)+\frac{7}{2}[/tex]
[tex]11=\frac{60}{8}+\frac{7}{2}[/tex]
[tex]11=7.5+3.5[/tex]
[tex]11=11[/tex]
This point is also on the line.
For (80,50):
[tex]50=\frac{5}{8}(80)+\frac{7}{2}[/tex]
[tex]50=50+3.5[/tex]
This point is not on the line, because it doesn't satisfy the linear equation.
For (-1, 2.875):
[tex]2.875=\frac{5}{8}(-1)+\frac{7}{2}[/tex]
[tex]2.875=\frac{5}{8}(-1)+\frac{7}{2}[/tex]
[tex]2.875=2.875/tex]
This also satisfy the equation.
Therefore, only one point is not on the line. All points on the line are:
(-1, 2.875); (12,11) and (0, 3.5).
The coordinate point (0 3.5) is on the line.
The coordinate point (12, 11) is on the line.
The coordinate point (80, 50) is NOT on the line.
The coordinate point (-1, 2.875) is on the line.
For us to determine whether the given points are not on the line, we need to get the required equation first.
The standard equation of a line is expressed as y-y0 = m(x-x0)
m is the slope
(x0, y0) is any point on the line:
[tex]m = \frac{6-1}{4+4}\\m=\frac{5}{8}[/tex]
Get the required equation:
[tex]y-1 = \frac{5}{8}(x+4)\\8y-8=5(x+4)\\8y-8=5x+20\\8y-5x=28[/tex]
For the point (0, 3.5)
When x = 0
8y - 5(0) = 28
8y = 28
y = 28/8
y = 3.5
This shows that the coordinate point (0, 3,5) is on the line.
For the point (12, 11)
When x = 12
8y - 5(12) = 28
8y - 60 = 28
8y = 88
y = 11
This shows that the coordinate point (12, 11) is on the line.
For the point (80, 50)
When x = 80
8y - 5(80) = 28
8y = 28 + 400
8y = 428
y = 428/8
y = 53.5
This shows that the coordinate point (80, 50) is NOT on the line.
For the point (-1, 2.875)
When x = -1
8y - 5(-1) = 28
8y + 5= 28
8y = 28 - 5
8y = 23
y = 2.875
This shows that the coordinate point (-1, 2.875) is on the line.
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