Find Possible Values of k in Co-ordinate Geometry Question

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Homework Help Overview

The discussion revolves around a coordinate geometry problem involving points A, B, and C with specific coordinates. The relationship between the distances AC and BC is defined, leading to a quadratic equation in terms of the variable k.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the Pythagorean theorem to derive equations for the distances AC and BC. There are attempts to simplify and solve a quadratic equation derived from these distances. Some participants express uncertainty about the correctness of their methods and results.

Discussion Status

There is an ongoing exploration of the problem, with participants sharing their equations and results. One participant acknowledges a mistake in their calculations, while another suggests a different approach that leads to a simpler quadratic equation. The discussion is productive, with various interpretations of the problem being examined.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific methods or formats for presenting their solutions. There is a focus on ensuring the relationships between distances are correctly established and interpreted.

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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
 
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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.
 
My equation becomes [tex]\[<br /> 0 = 76 + 68k + 15k^2 <br /> \][/tex]
 
... That suggests k = 2 or k= 2.53333333...repeating (two and eight-fifteenths)

..seems not to make sense.
 
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.
 
thnx, i made a careless error, lol
 

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