Question

A, B, and C are collinear, and B is between A and C. The ratio of AB to BC is 1:4.

Answer 1

Let point C has coordinates $(x_C,y_C).$ Consider vectors

$\overrightarrow{AB}=(x_B-x_A,y_B-y_A)=(8-9,6-9)=(-1,-3),\\ \\\overrightarrow{BC}=(x_C-x_B,y_C-y_B)=(x_C-8,y_C-6).$

Since the ratio AB to BC is 1:4, you have that

$\dfrac{-1}{x_C-8}=\dfrac{1}{4}\quad \text{and}\quad \dfrac{-3}{y_C-6}=\dfrac{1}{4}.$

Find $x_C$ and $y_C:$

$x_C-8=-4,\\ \\x_C=-4+8=4,\\ \\y_C-6=-3\cdot 4=-12,\\ \\y_C=-12+6=-6.$

Answer: C(4,-6)

Answer 2

Answer:

The coordinates of C are (4, -6)

Step-by-step explanation:

By the section formula,

The coordinate of a point that divides a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio of m : n are,

$(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$

Here,

$x_1=9, y_1=9, m = 1, n = 4$

Also,

$\frac{mx_2+nx_1}{m+n}=8$

$\frac{x_2+36}{1+4}=8$

$x_2+36=40$

$x_2=4$

$\frac{my_2+ny_1}{m+n}=6$

$\frac{y_2+36}{5}=6$

$y_2+36=30$

$y_2=-6$

Hence, the coordinates of C are (4, -6)