Locus and Roots/Coefficient Problem (1 Viewer)

[rama_v]

Hi Guys
I came across two problems I couldn't do, any help in solving them would be appreciated:

1) By using the perpendicular distance formula, find the locus of a point P(x,y) which is equidistant from the lines 3x + 4y =36 and 4x + 3y=24

2) If one zero of the function y=x^2 +mx +n is the square of the other, without finding the zeroes, prove that m^3 = n(3m - n -1)

Thanks

 

[FinalFantasy]

"2) If one zero of the function y=x^2 +mx +n is the square of the other, without finding the zeroes, prove that m^3 = n(3m - n -1)"
let zero's be A and B
A²=B
A+B=-m ------>(1)
AB=n--------->(2)

(1) A+A²=-m
(2) A³=n
-m³=(A(1+A))³=A³(1+3A+3A²+A³)=n(1+3A+3A²+n)=n(1+3(-m-B)+3(B)+n)
=n(1-3m-3B+3B+n)=n(1-3m+n)
therefore m³=-n(1-3m+n)=n(3m-n-1)

for da first one i forgot da formula, but just use it for each line and equate them den u should get da equation

 

[rama_v]

Thanks, I get it now

 

[FinalFantasy]

coool, btw wat was the formula u used for da first question lol

i seriously forgot it..

 

[rama_v]

Perpendicular Distance = absolute value of [ax1 + by1 +c1]/sqrt(a^2 +b^2)

 

[FinalFantasy]

ok thx
-----------------------

 

[Trev]

Haha, I couldn't remember it either when we had our mathematics exam, I was like errr..... meh. It doesn't count for me so I didn't care!

 

[davidgoes4wce]

Hi Guys
I came across two problems I couldn't do, any help in solving them would be appreciated:

1) By using the perpendicular distance formula, find the locus of a point P(x,y) which is equidistant from the lines 3x + 4y =36 and 4x + 3y=24

2) If one zero of the function y=x^2 +mx +n is the square of the other, without finding the zeroes, prove that m^3 = n(3m - n -1)

Thanks

Here is my solution to Q1.
































Part (b) asks us to draw 4 equations



From part (a) I assume that we only have to draw in the 2 equations instead of the 4, as 2 of them are identical? Could someone confirm.