Mathematics Question Thread

[bahramnilu]

Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7.


How would you do this question and why?

[8veFable]

Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7.


How would you do this question and why?

I’ll try my best but I’m not sure if I got the last part of this correct.

I would let P be point (x,y), and would use the perpendicular distance formula between P and each of those two lines, as the perpendicular formula allows you to find the shortest distance from a point to a line.

d = |a(x₁)+b(y₁)+c| / (a² + b²)^1/2

For P(x,y) and 4x-3y+2=0
|4(x) - 3(y) + 2|/√25     — (1)

For P(x,y) and 3x+4y-7=0
|3(x) + 4(y) - 7|/√25    — (2)

And since equations (1) and (2) give the distance from point P(x,y) to each line, and we want our point to be equistant (equal distance) from both of them, we will make them equal to each other.

|4x - 3y + 2|/√25 = |3x+4y-7|/√25

Cancel out the √25

|4x - 3y +2| = |3x + 4y - 7|

And then this is the dodgy part that I am not sure about. I would cancel out the absolute values but I have no idea if this is even allowed.

I got: x - 7y + 9 = 0, and graphing it on Desmos, it seems okay, but I would like to know if I broke mathematics with making the absolute values equal to each other and if it’s even allowed

[fun_jirachi]

The key point here is that its equidistant from both lines. This should probably give you the hint that you should be using the perpendicular distance formula on both lines, then equating them. In this case:


So we have two cases.
Case 1

Case 2


Notice that the two solutions are perpendicular to each other.

Hope this helps

EDIT: I saw your solution above, but you've got to note you can't just cancel the absolute values. When equating two absolute values, you have to consider both the negative case and the positive case. In a sense there are four cases, but you can narrow it down to two (which is what I've done) by considering that two pairs of cases are the same, where they are opposite in sign or have the same sign.

[8veFable]

Hi!

Thanks for fixing it up fun_jirachi 

I also have a small question that I wanted to get clarified about the 2016 HSC 2U Math question b) iii) --> Attached

For part iii) when you are expected to 'show' that dy/dt = y/400(200-y) I thought that you were not allowed to go 'reverse' and sub y into that equation and turn it into dy/dt but that is how the sample answers did it.

Is this a case with all math questions like this, where you are allowed to go 'reverse' and start with the equation that you are provided with to show that something is equal to something?

Thanks

[RuiAce]

Hi!

Thanks for fixing it up fun_jirachi 

I also have a small question that I wanted to get clarified about the 2016 HSC 2U Math question b) iii) --> Attached

For part iii) when you are expected to 'show' that dy/dt = y/400(200-y) I thought that you were not allowed to go 'reverse' and sub y into that equation and turn it into dy/dt but that is how the sample answers did it.

Is this a case with all math questions like this, where you are allowed to go 'reverse' and start with the equation that you are provided with to show that something is equal to something?

Thanks

You aren't working backwards there. You were given that \(y = \frac{200}{1+19e^{-0.5t}}\) at the start to begin with, and hence are not assuming anything you want to prove. Because you've already proven that, you can then sub it straight into \( \frac{y}{400}(200-y)\) and see what falls out.

What you want to prove is when you simplify that expression, coincidentally you get to the same answer as in (i). Which proves that in fact, \( \frac{y}{400}(200-y)\) is an alternate way of writing \( \frac{dy}{dt} \).

A case of working backwards would be assuming that \( \frac{dy}{dt} = \frac{y}{400}(200-y) \), and then subbing in your answer from part i), and rearranging the gibberish until you get something that's obviously true such as \(1 = 1\). Now that, would be unacceptable. Again, what you're doing here is manually computing \( \frac{y}{400}(200-y)\) given what you already know, and checking it actually equals \( \frac{dy}{dt} \).

Edit: Perhaps the resolve the confusion, a case of where you can't just sub \(y\) in without good reason would be if you weren't given the formula for \(y\) until that part. But here, we were given the formula for \(y\) at the very beginning.

[jamesrandom]

Hi, can someone please show me how to do this? I cant find which line of working i screwed up on. I can apply the quotient rule but I cant get the answer in the textbook which is

Spoiler

( x⁴ -2x³ -4x² -1 ) / ( x² - x - 1 )

[RuiAce]

Hi, can someone please show me how to do this? I cant find which line of working i screwed up on. I can apply the quotient rule but I cant get the answer in the textbook which is

Spoiler

( x⁴ -2x³ -4x² -1 ) / ( x² - x - 1 )

The denominator should have a square around it.
\begin{align*} \frac{d}{dx} \left( \frac{x^3+x}{x^2-x-1} \right)&= \frac{(3x^2+1)(x^2-x-1) - (2x-1)(x^3+x)}{(x^2-x-1)^2}\\ &= \frac{(3x^4-3x^3-3x^2+x^2-x-1)-(2x^4+2x^2-x^3-x)}{(x^2-x-1)^2}\\ &= \frac{x^4-2x^3-4x^2-1}{(x^2-x-1)^2}\end{align*}

[jamesrandom]

The denominator should have a square around it.
\begin{align*} \frac{d}{dx} \left( \frac{x^3+x}{x^2-x-1} \right)&= \frac{(3x^2+1)(x^2-x-1) - (2x-1)(x^3+x)}{(x^2-x-1)^2}\\ &= \frac{(3x^4-3x^3-3x^2+x^2-x-1)-(2x^4+2x^2-x^3-x)}{(x^2-x-1)^2}\\ &= \frac{x^4-2x^3-4x^2-1}{(x^2-x-1)^2}\end{align*}

Thanks mate, as I thought, I messed up on one line... and yes the denominator should be squared, i forgot to type that in my question.

[steviemay2000]

Could someone help clarify something about the attached question from the 2012 hsc exam. So I had a look at the success workbook solutions. And they substituted y into the formula and such to find the points of intersection and put the equation into a neat quadratic form. But they then went to say to let the discriminant be equal to zero since there are two equal roots...
I am confused by this, as how did they jump from finding the points of intersection to then using something to do with he discriminant?? I though the discriminant was to do with where the parabola cuts on the axis... Could someone please clarify my confusion. Thanks!

[RuiAce]

Could someone help clarify something about the attached question from the 2012 hsc exam. So I had a look at the success workbook solutions. And they substituted y into the formula and such to find the points of intersection and put the equation into a neat quadratic form. But they then went to say to let the discriminant be equal to zero since there are two equal roots...
I am confused by this, as how did they jump from finding the points of intersection to then using something to do with he discriminant?? I though the discriminant was to do with where the parabola cuts on the axis... Could someone please clarify my confusion. Thanks!

Actually, the discriminant being equal to zero is by definition what happens when there are two equal roots.

- \( \Delta > 0\) implies two distinct roots
- \( \Delta = 0\) implies two equal roots
- \( \Delta < 0\) implies no (real) roots

What you've stated is a follow-up result, and not a part of the definition of the discriminant. The idea is that if we then plot the graph of the quadratic (i.e. the new one we obtained from doing that subbing and rearranging), we get a (new) parabola. But, if the parabola has two equal roots, then the parabola barely touches the axis. Similarly, if the parabola has two distinct roots, then the parabola cuts the axis twice.

So, whilst the discriminant is related to where that new parabola cuts the axis, it's only related because it is a consequence of the three points I've stated above.

[parallaxd]

In 10aiv from the 2010 hsc maths exam I have no clue how to prove that result??

Thanks.

[RuiAce]

In 10aiv from the 2010 hsc maths exam I have no clue how to prove that result??

Thanks.

Simply use the range of the cosine function.

\( \cos \theta\) has range \(-1\leq \cos\theta \leq 1\).
Therefore \(-2\cos \theta\) has range \(-2\leq -2\cos\theta \leq 2\).
Therefore \(1-2\cos\theta\) has range \(-1\leq1-2\cos\theta\leq3\).

From here, we can just extract off the right bit to conclude that \(1-2\cos\theta\leq 3\). Multiplying both sides of the equation by \(a\) gives the desired result.

[Bells_123]

Hello! I was wondering with this multiple choice question, how to find all of the solutions of what x could be. I know the answer is C (6 solutions) but am not sure how to get all of them. Thank you!

[dazza2020]

Hello! I was wondering with this multiple choice question, how to find all of the solutions of what x could be. I know the answer is C (6 solutions) but am not sure how to get all of them. Thank you!

You'll have two equations to solve, sin3x = 1/2 & sin3x = -1/2, due to the nature of the absolute value
Then, as you're finding solutions for 3x, triple your range, and find all solutions for both equations for 0 < 3x < 3pi
FInally, divide all your solutions by 3 and you'll find 6 solutions betweem 0 < x < pi
The solutions are pi/18, 5pi/18, 13pi/18, 17pi/18, 7pi/18, 11pi/18

[8veFable]

Hi!

I'm having trouble with a line in the 2017 HSC Sample Answers for Q16 c) i) if anyone could please help me with it  - Attached

They get 'then BD = DE (equal intercept)', but I am unsure how they got this?

Much appreciated!!