# Homework Help: Co-ordinate geometry

1. Sep 17, 2007

### Trail_Builder

im stuck on an extension question to my hm/wk, hope you can help

1. The problem statement, all variables and given/known data

The points A, B and C have co-ordinate (-3, 2) (-1, -2), and (0, k) respectively, where k is constant.
Given that AC = 5BC, find the possible values of k.

2. Relevant equations

3. The attempt at a solution

I used pythagorus to get an equation for AC in terms of k, then the same for BC - then combined the 2 equations, simplified and ended up with 11 = 13k + 3k^2 then completed the square ending up with k = +-(squareroot)(301/36) - (13/6)

some how I think I've gone wrong or got the wrong method.

can someone please guide me :D thnx

2. Sep 17, 2007

### Dick

The method seems to be right. The answer is wrong. You'll need to show us the details of how you got the quadratic to solve.

3. Sep 17, 2007

### symbolipoint

My equation becomes $$$0 = 76 + 68k + 15k^2$$$

4. Sep 17, 2007

### symbolipoint

... That suggests k = 2 or k= 2.53333333...repeating (two and eight-fifteenths)

..seems not to make sense.

5. Sep 17, 2007

### HallsofIvy

The distance from A to C, squared, is (-3)2+ (2-k)2= 9+ 4- 4k+ k2. The distance from B to C, squared is (-1)2+ (-2-k)2= 1+ 4+ 4k+ k2. Saying that AC= 5BC is the same as AC2= 25BC2 or 13- 4k+ k2= 25(5+ 4k+ k2)= 125+ 100k+ 25k2. That gives 24k2+ 104k+ 112= 0 Dividing through by 8, that is 3k2+ 13k+ 14= 0. That doesn't look like what you got! And it factors rather easily! That's always a good sign.

6. Sep 19, 2007

### Trail_Builder

thnx, i made a careless error, lol