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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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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.`
Posted by sciencesolve on August 12, 2013 at 4:17 PM (Answer #1)
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