# Given A(1,2), B(3,1), what is real m if mid point of AB belong to x+y-m=0?

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### 1 Answer

You need to remember what is the condition for a point to lie on a line, hence, you need to test if the equation of the line holds for the values of coordinates of the point.

Hence, you need to evaluate the coordinates of the middle point M of the line segment AB, thus, you need to use the midpoint formula, such that:

`x_M = (x_A + x_B)/2 => x_M = (1 + 3)/2 => x_M = 2`

`y_M = (y_A + y_B)/2 => y_M = (1 + 2)/2 => y_M = 3/2`

Since the problem provides the information that the middle point lies on the line `x + y - m = 0` , you need to replace `x_M` for x and `y_M` for y in equation, such that:

`x_M + y_M - m = 0 => 2 + 3/2 - m = 0`

Isolating m to the left side, yields:

`-m = -2 - 3/2 => m = 2 + 3/2 => m = 7/2`

**Hence, evaluating m, under the given conditions, yields **`m = 7/2.`