VCE Methods Question Thread!

[Shadowxo]

http://m.imgur.com/a/O3hHR

I'm abit stuck on this question. I know that I need to find the gradients of the line but I'm having trouble finding the point of intersection which is the last bit of info I need.

Thanks

Hi, you're almost there. You've found the possible values of k, and k is the gradient of the line. The line has the equation y = kx + b so just substitute each value of k in to find the two equations, and as y=3 when x=0, b = 3 and substitute those values into k
Should end up with y = 3x + 3 and y = -x + 3

[Alicia23]

Hi guys, I'm stuck on this question,
"Prove that the equation x^2+(a+1)x+(a-2)=0 always have two distinct solutions."
The answer just states "show that the discriminant > 0 for all a"
Thanks heaps

[Escobar]

Hi guys, I'm stuck on this question,
"Prove that the equation x^2+(a+1)x+(a-2)=0 always have two distinct solutions."
The answer just states "show that the discriminant > 0 for all a"
Thanks heaps

for an equation of the form ax^2+bx+c, the discriminant is b^2-4ac
when discriminant>0, there are 2 real solutions
when discriminant = 0, there is 1 real solution
when discriminant<0, there are no real solutions

in this question, the discriminant is (a+1)^2-4(a-2)
=a^2+2a+1-4a+8
=a^2+2a-4a+9
=a^2-2a+9
=(a-1)^2+8 by completing the square
(a-1)^2>0 and 8>0 so (a-1)^2+8>0
so discriminant>0

[cutiepie30]

How would you do these questions ?

[Escobar]

How would you do these questions ?

1st question

2nd question

[cutiepie30]

1st question

2nd question

Thanks Escobar 

[Guideme]

Help pls! Thank you!

[Shadowxo]

(Image removed from quote.)
Help pls! Thank you!

Hi,
So the area of the rectangle is length x width, let's call this x and y. A = xy
The radius is a, so if we have a right angled triangle dividing the rectangle in half, 2a is the hypotenuse.
x² + y² = (2a)²
y = √(4a²-x²)
Area = xy = x√(4a²-x²)

Domain: x must be greater than 0 (so it has an length and area) and less than 2a (so there is a width and area)
Domain = (0,2a)

Does this help?

[geminii]

Hi everyone,

I've been trying to do this seemingly simple question and I can't seem to figure it out.

The question is:

The length of the line segment joining A(2,6) and B(10,y) is 10 units. Find y.

This is my working:

10 = √[(10-2)^2 + (y-6)^2]
10 = √[64 + (y-6)^2]
10 = 8 + √[(y-6^2]
10 = 8 + (y-6)
2 + 6 = y
y = 8

However, the answer in the textbook says y is meant to equal 12 or 0.

I used the following formula in my working:
AB = √[(x2-x1)^2 + (y2-y1)^2]

Could anyone please help? Thanks in advance!!

[Shadowxo]

Hi everyone,

I've been trying to do this seemingly simple question and I can't seem to figure it out.

The question is:

The length of the line segment joining A(2,6) and B(10,y) is 10 units. Find y.

This is my working:

10 = √[(10-2)^2 + (y-6)^2]
10 = √[64 + (y-6)^2]
10 = 8 + √[(y-6^2]
10 = 8 + (y-6)
2 + 6 = y
y = 8

However, the answer in the textbook says y is meant to equal 12 or 0.

I used the following formula in my working:
AB = √[(x2-x1)^2 + (y2-y1)^2]

Could anyone please help? Thanks in advance!!

Hi! just a little mistake
√(a+b) doesn't equal √a + √b
to figure out y, you need to square both sides
100 = 64 + (y-6)²
(y-6)² = 36
y-6 = +-6
y = 0 or 12

[geminii]

Hi! just a little mistake
√(a+b) doesn't equal √a + √b
to figure out y, you need to square both sides
100 = 64 + (y-6)²
(y-6)² = 36
y-6 = +-6
y = 0 or 12

Ah, thank you so much!!

[Shadowxo]

 No worries

[geminii]

I have another question -

Find the angle that the line joining the given points makes with the positive direction of the x axis.
a) (-4,1), (4,6)

The textbook says to use the formula m = tan(theta)
where m is the gradient.
I got 5/8 as the gradient (rise over run using the x and y coordinates).

But when I do this in the following modes on the calculator, I don't get the correct answer of 32.01 degrees:
Decimal, Real, Gra: 0.625=tan(1.1111111111*theta)
Decimal, Real, Rad: 0.625=tan(0.017452....*theta)
Decimal, Real, Deg: 0.625=tan(theta)

Am I missing something?

[Shadowxo]

I have another question -

Find the angle that the line joining the given points makes with the positive direction of the x axis.
a) (-4,1), (4,6)

The textbook says to use the formula m = tan(theta)
where m is the gradient.

But when I do this in the following modes on the calculator, I don't get the correct answer of 32.01 degrees:
Decimal, Real, Gra: 0.625=tan(1.1111111111*theta)
Decimal, Real, Rad: 0.625=tan(0.017452....*theta)
Decimal, Real, Deg: 0.625=tan(theta)

Am I missing something?

I'm not sure what you're doing but I always do the inverse to find the angle (and have it set to degrees for questions like this)
tanx = 5/8
tan^-1(5/8) = x (should be a tan^-1 button on calc)
x = 32º

[geminii]

I'm not sure what you're doing but I always do the inverse to find the angle (and have it set to degrees for questions like this)
tanx = 5/8
tan^-1(5/8) = x (should be a tan^-1 button on calc)
x = 32º

Oh ok thank you! I was typing 5/8 = tan(theta) directly into the calculator.