# colinear problempoints A(8,0),B(3,6),C(0,3) BC intersect 0x in D AB intersect 0y in E prove the midpoints of OB,AC,DE are colinear

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### 1 Answer

You need to find the equation of line `BC` such that:

`y - y_C = (y_B - y_C)/(x_B - x_C)(x - x_C)`

`y - 3 = (6 - 3)/(3 - 0)(x - 0) => y - 3 = x => y = x + 3`

Since the problem provides the information that the line `BC` crosses x axis in the point D, you may find the coordinates of the point D such that:

`0 = x + 3 => x = -3`

Hence, the line `BC` intersect x axis at `D(-3,0).`

You need to find the equation of line `AB ` such that:

`y - y_B = (y_A - y_B)/(x_A - x_B)(x - x_B)`

`y -6 = (0 - 6)/(8 - 3)(x - 3) => y - 6 = -6/5(x - 3)`

Since the problem provides the information that the line `AB` crosses y axis in the point E, you may find the coordinates of the point E such that:

`y - 6 = (-6/5)(0 - 3) => y = 6 + 18/5 => y = 48/5`

Hence, the line `AB` intersect y axis at `E(0,48/5).`

You need to find the midpoint `OB` such that:

`x_1 =(x_O + x_B)/2 => x_1 = (0 + 3)/2 = 3/2`

`y_1 = (y_O + y_B)/2 => y_1 = (0 + 6)/2 = 3`

You need to find the midpoint AC such that:

`x_2 = (x_A + x_C)/2 => x_2 = (8 + 0)/2 = 4`

`y_2 = (y_A + y_C)/2 => y_2 = (0 + 3)/2 = 3/2`

You need to find the midpoint DE such that:

`x_3 = (x_D + x_E)/2 => x_3 = (-3 + 0)/2 = -3/2`

`y_3 = (y_D + y_E)/2 => y_3 = (0 + 48/5)/2 = 24/5`

You need to check if the midpoints are collinear evaluating the following determinant, such that:

`[(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)] = [(3/2,3,1),(4,3/2,1),(-3/2,24/5,1)]` `= 9/4 + 96/5 - 9/2 + 9/4 - 72/10 - 12 = 96/5 - 72/10 - 12 = (192 - 72 - 120)/10 = (192 - 192)/10 = 0` **Hence, since evaluating the determinant of the midpoints yields 0, hence, the midpoints are collinear.**