1995 maths leaving cert paper 2

alic

just doing the questions and want to check some of my work any idea where i can get the answers off this cant find a website that goes back to 1995 there no e-xamit code with it either

the question is on co-ordinate geometry,paper 2 qs 2

ryanairzer

The answers are at the back of the EDCO exam papers.

alic

i know but i need how they are done it doesnt tell how the question is done in the back its only one question anyway

part c:

c(1,6) and d(-3,-1) are two points. The point r has coordinates (2,y) such that |cd| = |cr|

Find the two possible vales of y

ryanairzer

c(1,6) and d(-3,-1) are two points. The point r has coordinates (2,y) such that |cd| = |cr|

Okay we need to use our distance formula for this.

What you do is find the distance from c to d using the distance formula.

Then, find the distance in terms of y from c to r using the distance formula.

Then let your first distance equal your second distance. Bring everything to one side and you've got a quadratic equation in terms of y, which you then solve.

timmywex

ryanairzer wrote: »

c(1,6) and d(-3,-1) are two points. The point r has coordinates (2,y) such that |cd| = |cr|

Okay we need to use our distance formula for this.

What you do is find the distance from c to d using the distance formula.

Then, find the distance in terms of y from c to r using the distance formula.

Then let your first distance equal your second distance. Bring everything to one side and you've got a quadratic equation in terms of y, which you then solve.

yeh
*=root

*65=*y^2-4y+5.................these are results for after doing distance formulae
65=y^2-4y+5
y^2-4y-60=0
y=10 or y=-6
y=-10 or y=6

that should be right, i trust you are able to do the distance formulae yourself

alic

timmywex wrote: »

yeh
*=root

*65=*y^2-4y+5.................these are results for after doing distance formulae
65=y^2-4y+5
y^2-4y-60=0
y=10 or y=-6
y=-10 or y=6

that should be right, i trust you are able to do the distance formulae yourself

no the answer is 14 or - 2 but thanks for the help

ZorbaTehZ

|cd|=|cr|
√(-3-1)^2+(-1-6)^2=√(2-1)^2+(y-6)^2
16+49=1+y^2-12y+36
65=y^2-12y+37
y^2-12y-28=0
(y+2)(y-14)=0
y=-2, y=14

alic

i dont understand where you got that 12y

but thanks been at me all day very annoying when your good at something then get a qs like that

Fringe

(y - 6)^2
(y- 6)(y - 6)
y^2 - 6y - 6y + 36
y^2 - 12y + 36

timmywex

alic wrote: »

no the answer is 14 or - 2 but thanks for the help

oh sorry, i took the wrong two points.