Help for Maths Question! (1 Viewer)

[lookoutastroboy]

hey guys,

there's a maths problem in my book that I can't do since its so infuriatingly hard and has taken me ages to complete, if you can give me a solution and complete working out that'd be great,

well, here it is:

In the diagram, the line 5x + 2y + 5 = 0 cuts the x-axis and y-axis at E and C respectively. BD is the line x = 2, AB is parallel to the x-axis and BE and CD are perpendicular to AC. Find the co-ordinates of the points A, B,C,D,E.


View attachment 16470

 

[bored of sc]

hey guys,

there's a maths problem in my book that I can't do since its so infuriatingly hard and has taken me ages to complete, if you can give me a solution and complete working out that'd be great,

well, here it is:

In the diagram, the line 5x + 2y + 5 = 0 cuts the x-axis and y-axis at E and C respectively. BD is the line x = 2, AB is parallel to the x-axis and BE and CD are perpendicular to AC. Find the co-ordinates of the points A, B,C,D,E.


View attachment 16470

E = (-1,0) (sub y = 0 into equation since E lies on the x-axis meaning its y coordinate is 0).

C = (0, -5/2) (as above but sub in x = 0)

As for the rest, you need negative reciprocle (for perpendicular) of the line but I forget how to do it, sorry...

On second attempt with help of lyounamu they are the correct answers .

 

[lyounamu]

E = -2.5 (sub x = 0 into equation since E lies on the x-axis meaning its x coordinate is 0).

C = -1 (as above but sub in y = 0)

As for the rest, you need negative reciprocle (for perpendicular) of the line but I forget how to do it, sorry...

I think the above two coordinates are correct but check with someone else better at maths than me.

I cannot say that I am better but the above answers are wrong.

C is a y-intercept and E is a x-intercept so:

C: (0, -5/2)
E: (-1,0)

EDIT: I will post up answers ASAP

 

[bored of sc]

I cannot say that I am better but the above answers are wrong.

C is a y-intercept and E is a x-intercept so:

C: (0, -5/2)
E: (-1,0)

oh yer, true... my bad

and just for the record, you are WAY better at maths than me - obviously

 

[lyounamu]

hey guys,

there's a maths problem in my book that I can't do since its so infuriatingly hard and has taken me ages to complete, if you can give me a solution and complete working out that'd be great,

well, here it is:

In the diagram, the line 5x + 2y + 5 = 0 cuts the x-axis and y-axis at E and C respectively. BD is the line x = 2, AB is parallel to the x-axis and BE and CD are perpendicular to AC. Find the co-ordinates of the points A, B,C,D,E.


View attachment 16470

You know how I found E and C, right?

You have to find B and D first. So, since the gradient of CE is -5/2, the gradient of BE and CD are 2/5

So, here you use the fact that the gradient of BE is 2/5 to find that B is (2, 6/5)

Then, you can also use the fact that the gradient fof CD is 2/5 to find that D is (2, -17/10)

To find A, just write that gradient of BE = -1/gradient of AE
Therefore, A: (-37/25, 6/5)

This is a poor working out. I will write it down and post it up by tomorrow (by camera shot). Sorry about this pathetic working out.

 

[lookoutastroboy]

thanks lyounamu,


i understand the maths problem now!!!

 

[lookoutastroboy]

anyways,


if ya wanna supply that camera shot, that'd be appreciated too.










thanks, lookoutastroboy.

 

[lyounamu]

I gotta admit: it was the hardest yr 10 question that I have ever seen. I was quite relaxed when I first attempted the question but after the first attempt, I knew that this question was a serious business. I will post up my detailed working out ASAP. I just don't know how to post this complicated one up.

 

[tommykins]

No need lyou - I've got it.

http://img183.imageshack.us/img183/508/72210387yl4.jpg

A(37/25, 6/5)
B(2,6/5)
C(0,-5/2)
D(2,-17/10)
E(-1,0)

I will go out of order since it's easier in the end.

For D

The gradient of 5x+2y+5 = 0 is -5/2 (rearranging it).

Since CD is perpindicular to AC, the gradient of CD becomes 2/5.
C has co-ordinates (0,-5/2)

CD is y+5/2 = 2/5(x-0) which works out to be 10y + 25 = 4x

Since D is on the line x = 2, sub that into this equation.
10y+25 = 8 leading to y = -17/10

Thus D is (2,-17/10)

For B
The gradient of EB wil be 2/5 (since once again, it's perpindicular to AC).
E has co-ordinates (-1,0)
y - 0 = 2/5(x+1) which leads to 5y = 2x+2

B is on the line x =2, sub it in

5y = 4+2 leading to y = 6/5

Therefore, B is (2,6/5)

For A

Since B is (2,6/5) and AB is a straight horizontal line parallel to the x axis, it's gradient will be 0.

AB will then be y- 6/5 = 0(x-2) thus the equation of the line AB is y = 6/5

Subbing y = 6/5 into 5x+2y+5 = 0 (since A is on this line as well)
5x + 12/5 + 5 = 0
25x + 12 + 25 = 0

.:. x = -37/25

Thus, A is (-37/25,6/5)


E and C are axis so you just sub x = 0 and y = 0 into 5x+2y + 5 = 0 to get their coordinates.

This question is within the scope of a year 10 student, but figuring it out and it's tedious algebra (in terms of the year 10 student) makes any year 10 student who is able to do this, really bright at mathematics.

If you have any questions, simply post here.

 

[lyounamu]

No need lyou - I've got it.

http://img183.imageshack.us/img183/5661/70822736zl0.jpg

A(37/25, 6/5)
B(2,6/5)
C(0,-5/2)
D(2,-17/10)
E(-1,0)

I will go out of order since it's easier in the end.

For D

The gradient of 5x+2y+5 = 0 is -5/2 (rearranging it).

Since CD is perpindicular to AC, the gradient of CD becomes 2/5.
C has co-ordinates (0,-5/2)

CD is y+5/2 = 2/5(x-0) which works out to be 10y + 25 = 4x

Since D is on the line x = 2, sub that into this equation.
10y+25 = 8 leading to y = -17/10

Thus D is (2,-17/10)

For B
The gradient of EB wil be 2/5 (since once again, it's perpindicular to AC).
E has co-ordinates (-1,0)
y - 0 = 2/5(x+1) which leads to 5y = 2x+2

B is on the line x =2, sub it in

5y = 4+2 leading to y = 6/5

Therefore, B is (2,6/5)

For A

Since B is (2,6/5) and AB is a straight horizontal line parallel to the x axis, it's gradient will be 0.

AB will then be y- 6/5 = 0(x-2) thus the equation of the line AB is y = 6/5

Subbing y = 6/5 into 5x+2y+5 = 0 (since A is on this line as well)
5x + 12/5 + 5 = 0
25x + 12 + 25 = 0

.:. x = 37/25

Thus, A is (37/25,6/5)


If you have any questions, simply post here.

WTH! I want my recognition!!! I got my answers right!!!

By the way, your answer for A is wrongly written. It is -37/25 not 37/25. Any recognition for this???

EDIT: I really thank you for this. I was just joking above. Thank you for clarification & perfect working out.

 

[tommykins]

WTH! I want my recognition!!! I got my answers right!!!

By the way, your answer for A is wrongly written. It is -37/25 not 37/25. Any recognition for this???

EDIT: I really thank you for this. I was just joking above. Thank you for clarification & perfect working out.

Yea, it's -37/25, thought I typed it out.

Saved you the trouble of typing this out, I actually wrote it on paper but the printer sucks at scanning.