Which is the point that has similar coordinates and it is on the line y=(x-1)/2?

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Testing the point coordinates in equation of line yields:

`y_A = (x_A - 1)/2`

The problem provides the information that `x_A = y_A` such that:

`x_A = (x_A - 1)/2 => x_A - (x_A - 1)/2 = 0`

`x_A - x_A/2 + 1/2 = 0 => x_A - x_A/2 = - 1/2 => x_A/2 = -1/2 `

Reducing duplicate factors yields:

`{(x_A = -1),(x_A = y_A):} => y_A = -1`

**Hence, evaluating the missing coordinates, under the given conditions, yields **`x_A = -1, y_A = -1.`

We'll note the point that has like coordinates as M(m,m).

Since the point is located on the line y = (x - 1)/2, it's coordinates verify the expression of the line.

We'll put y = f(x) and we'll substitute x and y by the coordinates of the given point:

f(m) = (m - 1)/2 (1)

But f(m) = m (2)

We'll conclude from (1) and (2) that:

(m - 1)/2 = m

We'll cross multiply and we'll get:

2m = m - 1

We'll isolate m to the left side. For this reason, we'll subtract m both sides:

2m - m = -1

m = -1

The point located on the line y = (x-1)/2 has the coordinates (-1;-1).

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