A(-3, 4, -7), B(5, 5, 5) and C(7, 9, 8) are collinear - Scottish Highers Maths - Question 5 - 2018

To find the ratio in which point B divides line segment AC, we can use the section formula. The coordinates of point A are (-3, 4, -7), B are (5, 5, 5), and C are (7, 9, 8).

Using the formula for the ratio: If a point divides the line segment joining points A(x1, y1, z1) and C(x2, y2, z2) in the ratio m:n, then:

$B = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$

Substituting the coordinates into the formula, we have:

$B = (5, 5, 5) \text{ and } A = (-3, 4, -7), C = (7, 9, 8)$

We can first set up equations for x, y, and z coordinates, which should give us the ratios based on the components:

1. For x-coordinates: $5 = \frac{m(7) + n(-3)}{m+n}$

2. For y-coordinates: $5 = \frac{m(9) + n(4)}{m+n}$

3. For z-coordinates: $5 = \frac{m(8) + n(-7)}{m+n}$

From these equations, we find that the x-coordinates give us the ratio 1:4. Therefore, B divides AC in the ratio of 1:4.