A, B, and C are collinear, and B is between A and C. The ratio of AB to AC is 1:3. If A is at (-2,-3) and B is at (-6,-1), what are the coordinates of point C

Answer 1

If the point partitions the line internally then the formula used is: ( (m x2 + n x1) / (m+n) , (m y2 + n y1) / (m+n) ), If we consider m : n is the ratio and point A as (x1 , y1) and point B as (x2, y2).

If the point partitions the line externally then the formula used is: ( (m x2 - n x1) / (m-n) , (m y2 - n y1) / (m-n) ), If we consider m : n is the ratio and point A as (x1 , y1) and point B as (x2, y2).

Coming to the question, A ,B and C are collinear, B is between A and C, AB : AC is given and we have to find point C.

So, C partitions the line AB externally.

C(x3 , y3) = ( (m x2 - n x1) / (m-n) , (m y2 - n y1) / (m-n) ).

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Answer 2

between points A and B:

Δx = x_B - x_A = (-6) - (-2) = -4

Δy = y_B - y_A = (-1) - (-3) = 2

Since the ratio of AB to AC is 1:3, you can multiply the changes in x and y by 3 and add that to the x and y coordinates of point B to find point C.

between points B and C:

Δx = -4 * 3 = -12 → x_C = x_B + Δx = -6 + (-12) = -18

Δy = 2 * 3 = 6 → y_C = y_B + Δy = -1 + 6 = 5

point C = (-18,5)