Find Possible Values of k in Co-ordinate Geometry Question

[Trail_Builder]

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

Homework Statement

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.

Homework Equations

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

 

[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.

 

[symbolipoint]

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

 

[symbolipoint]

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

..seems not to make sense.

 

[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.

 

[Trail_Builder]

thnx, i made a careless error, lol