Question

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

If A is at (9,9) and B is at (8,6), what are the coordinates of point C?

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Answer 1

The coordinates of C are (4, -6)

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,

`((mx_2+nx_1)/(m+n), (my_2+ny_1)/(m+n))`

Here,

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

Also,

`(mx_2+nx_1)/(m+n)=8`

`(x_2+36)/(1+4)=8`

`x_2+36=40`

`x_2=4`

`(my_2+ny_1)/(m+n)=6`

`(y_2+36)/(5)=6`

`y_2+36=30`

`y_2=-6`

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

Answer 2

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

`(-1)/(x_C-8)=(1)/(4)\quad \text{and}\quad (-3)/(y_C-6)=(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)