If the coordinates 3,7 -6,1 and 9,p lie on the same line find the value of p and explain your steps 

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If three points lie on the same line, which means collinear, the area of the triangle formed by them is 0.

Let us substitute these points in the formula of the area of a triangle, i.e.

`1/2 (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) = 0`

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If three points lie on the same line, which means collinear, the area of the triangle formed by them is 0.

Let us substitute these points in the formula of the area of a triangle, i.e.

`1/2 (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) = 0`

`(x1, y1) = (3,7)`

`(x2, y2) = (-6,1)`

`(x3, y3) = (9,p)`

`1/2(3(1-p) + (-6)(p-7) + 9(7-1)) = 0`

`3 - 3p - 6p + 42 + 54 = 0`

`-9p + 99 = 0`

`p = 99/9 = 11`

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Hello!

Let's write an equation of the straight line which goes through the first two points. Then the third point lies on this line if and only if its coordinates satisfy the equation.

For two points (x1, y1) and (x2, y2) the equation of the line is

`(x-x1)/(x2-x1)=(y-y1)/(y2-y1).`

Here the specific equation is

`(x-3)/(-6-3)=(y-7)/(1-7),`

or

`(x-3)/(-9)=(y-7)/(-6),`

or

`6*(x-3)=9*(y-7).`

Which simplifies to: 

`y = (2/3)x + 5`

Now substitute x=9 and y=p (the coordinates of the third point) into this equation:

`p = (2/3) * 9 + 5 `

or

p=11. This is the answer.

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