Question

ABC is a straight line such that BC=2AB. If the coordinates of A are (1,-5) and the coordinates of B are (6,2),find the coordinates of C?

Answer 1

Try this:

Explanation:

Considering the diagram:

The distance `AB` is:
`d_(AB)=sqrt((x_B-x_A)^2-(y_B-y_A)^2)=sqrt((6-1)^2+(2+5)^2)=sqrt(25+49)=8.6`
But:
`d_(BC)=2d_(AC)=2*8.6=17.2`
So:
`d_(BC)=17.2=sqrt((x_C-x_B)^2-(y_C-y_B)^2)`

I found that the equation of the line through A and B is:
`(x-x_B)/(x_B-x_A)=(y-y_B)/(y_B-y_A)`
`(x-6)/(6-1)=(y-2)/(2+5)`
`y=7/5x-32/5`
so that:
`y_C=7/5x_C-32/5`

substituting into:
`d_(BC)=17.2=sqrt((x_C-x_B)^2-(y_C-y_B)^2)`

I get:
`17.2=sqrt((x_C-6)^2-(7/5x_C-32/5-2)^2)`
square both sides:
`(17.2)^2=(x_C-6)^2-(7/5x_C-32/5-2)^2`
`295.84=x_C^2-12x_C+36+49/25x_C^2-588/25x_C+1764/25`

rearranging you get:
`74x_C^2:-888x_C-4732=0`

Solving using the Quadratic Formula you get:
`x_(C1,2)=(888+-sqrt(788544+1400672))/148`
So:
`x_(C1)=16`
`x_(C2)=-4`

Use `x_(C1)=16` into `y_C=7/5x_C-32/5` you get `y_C=16`

Try the other value of `x_(C2)` to get the other answer.