Question

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

Answer 1

Answer:

The coordinates of C is (-8,1)

Step-by-step explanation:

Looking at the attached image, you'd see that, AB is 1 and BC is 1, that's because we are told that ratio of AB to AC is 1:2 meaning, AC is 2 and AB is 1. Therefore for that ratio to be satisfied BC has to be 1 so that AC would be 2.

Now let's assume the coordinates of C is $(x,y)$.

To get it's coordinates, we use the section formula:

$(x,y)= (\frac{mx + nx_{1}}{m + n} , \frac{my + ny_{1}}{m + n})$

Where (m,n) is the (AB, BC)

Therefore we have

$(-1,2) = (\frac{1\times x + 1\times6}{1 + 1} , \frac{1\times y + 1\times3}{1 + 1})$

This gives:

$(-1,2) = (\frac{x + 6}{2} , \frac{y + 3}{2})$

$-1 =( \frac{x + 6}{2}$ and $2 = \frac{y + 3}{2})$

From there x = -8 and y = 1

Therefore the coordinates of C is (-8, 1).