What is the point C that belong to x+y-1=0 if C is colinear to A(1,2), B(3,1)?

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You should come up with the following notation for coordinates of the point C, such that:

`x_C = a, y_C = b`

Since the problem provides the information that the point C belongs to the line `x + y = 1` , hence, you may replace a for x and b for y in equation of line, such that:

`a + b = 1 => a = 1 - b`

You should use the following determinant condition for three collinear points, such that:

`[(x_A,y_A,1),(x_B,y_B,1),(x_C,y_C,1)] = 0`

`[(1,2,1),(3,1,1),(1 - b,b,1)] = 0``=> 1 + 3b + 2(1-b) - (1-b) - b - 6 = 0`

`1 + 3b + 2 - 2b - 1 + b - b - 6 = 0`

Reducing duplicate members and isolating the term that contains b to the left side, yields:

`b = 4 => a = 1 - 4 => a = -3`

**Hence, evaluating the coordinates of the point C, under the given conditions, yields **`x_C = -3, y_C = 4.`

**Sources:**

Find the point C that lies on the line x+y-1=0 and the line connecting A(1,2) and B(3,1):

The line through A and B: the slope is `m=(1-2)/(3-1)=-1/2` .

Use the point-slope form to find the equation:

`y-2=-1/2(x-1)==> y=-1/2x+5/2`

Write the other equation is slope-intercept form `y=-x+1`

Since C lies on both lines,the y-values are equal:

`-1/2 x+5/2=-x+1`

`1/2x=-3/2`

`x=-3==> y=4`

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The point C is (-3,4)

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The graphs:

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