Coordinate Geometry: Determination of Section Ratio

Determining the Section Ratio

The coordinate of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are given by
x = (mx2 + nx1)/ (m + n), y = (my2 + ny1) / (m + n)

Example: Find the coordinates of the point which divides the line segment joining the points (1, -2) and (-3, 4) in the ratio 5 : 3 internally.
 Solution: Let P (x, y) be the required point. Then, x = [5 * (-3) + 3 * 1]/(5 + 3) and y = [5 * 4 + 3 * (-2)]/(5 + 3) x = [(-15) + 3]/8 and y = [20 + (-6)]/8 x = -3/2 and y = 7/4
So, the coordinates of point P are (-3/2, 7/4).

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 Example: Determine the ratio in which the line 3x + y - 9 = 0 divides the segment joining the points (2, 5) and (4, 9). Solution: Suppose the line 3x + y - 9 =0 divides the line segment joining A(2, 5) and B(4, 9) in the ratio k : 1 at point P. Then, the coordinates of P are [(4k + 2)/(k + 1), (9k + 5)/(k + 1)] But, P lies on 3x + y - 9 = 0. Therefore, [3(4k + 2)/(k + 1)] + [(9k + 5)/(k + 1)] - 9 = 0 12k + 6 + 9k + 5 - 9k - 9 = 0 12k - 2 = 0 k = 1/6 So, the required ratio is 1 : 6 internally.

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