[luke]

Tried to draw this on paint I don't think it went too badly. Sorry for the inaccuracy.
I'm completely stuck on this and have NO idea what to do after these steps.

[dugdugdug]

(Original post by luke)


Tried to draw this on paint I don't think it went too badly. Sorry for the inaccuracy.
I'm completely stuck on this and have NO idea what to do after these steps.

Try solving the following simultaneous equations:

(5-y)^2+(4-x)^2=5 and x^2+(7-y)^2=25

[Fortune_3]

(Original post by luke)


Tried to draw this on paint I don't think it went too badly. Sorry for the inaccuracy.
I'm completely stuck on this and have NO idea what to do after these steps.

Okay, so you know that the equation of DE is y=-1/2x + 7, so therefore the equation of EF is y=2x-3 (using perpendicular lines, and the point (4,5), which is on the line EF

You are told that the diameter is 5 units, and you know you have a right angle triangle. You can work out the length from D to E, which is 2root5, so using Pythagoras you know that the distance from E to F is root 5 (apprx 2 down). So you know that your y coordinate is 3 (because you have (4,5), and 2 down from this y coordinate is 3)

Therefore, you can sub in this stuff into the formula y=2x-3, to get an x coordinate of 3.

Point F is (3,3).... I think

[luke]

Thanks both of you the solution I came up with eventually is that if DF is 5 units, the locus of points that are 5 units away from D is x^2 + (y-7)^2 = 25, as the equation of Line EF is y=2x-3. Point F satisfies both of these so essentially you solve the simultaneous equations of which are x=3 or x=5, as 0<x<4 then x = 3 and then plug that back in to find out y = 3 also.

So F is (3,3)

[dugdugdug]

(Original post by luke)
Thanks both of you the solution I came up with eventually is that if DF is 5 units, the locus of points that are 5 units away from D is x^2 + (y-7)^2 = 25, as the equation of Line EF is y=2x-3. Point F satisfies both of these so essentially you solve the simultaneous equations of which are x=3 or x=5, as 0<x<4 then x = 3 and then plug that back in to find out y = 3 also.

So F is (3,3)

I'm glad you got the correct answer (3,3) which "proved" my simultaneous equations were correct.

Whilst attempting to solve the problem, I drew a number of right angled triangles to get to my two equations. In these questions, it helps to know how many marks were at stake because if it's only say 3-4 marks, I'm convinced I took a long winded way to get at the answer.

[luke]

(Original post by dugdugdug)
I'm glad you got the correct answer (3,3) which "proved" my simultaneous equations were correct.

Whilst attempting to solve the problem, I drew a number of right angled triangles to get to my two equations. In these questions, it helps to know how many marks were at stake because if it's only say 3-4 marks, I'm convinced I took a long winded way to get at the answer.

Most of the questions were 3-5 marks...however this one was worth 8....