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[aequilibrium]

@SilverIcyCat

x and y in your formula are the variables abscissa and ordinate, that define the coordinates of each point in your line. I'm not sure if this is what you wanted as an answer.

So, moving on to the resolution.


a)
f the triangle is right angled at A, it means [AB] and [AC] are perpendicular to each other, so you can use the vectorial scalar product (?) - I don't know what it's called in english, sorry.

1. Define the two vectors:

AB = B - A = (1, 3) - (-2, -3) = (3, 6)
AC = C - A = (10,K) - (-2, -3) = (12, k+3)

Note: don't forget you have to add the arrow on top of the AB and AC since they are vectors.


2. Use the formula and proprieties:

You know the scalar product is 0 when two vectors are perpendicular, which means AB . AC = 0

AB . AC = 0
(3, 6) . (12, k+3) = 0
3 x 12 + 6.(k+3) = 0
36 + 6k + 18 = 0
6k = -54
k = -9

Now you know that C (10, -9) and you have the triangle defined.





b)
1. Calculate M:

M is midway in the segment [BC], so all you have to do is divide the length in half. So, considering B (b1, b2), C (c1, c2) and M (m1, m2), the coordinates of M can be calculated like this:

x = (b1 + c1) / 2
y = (b2 + c2) / 2

x = (1 + 10) / 2 = 11/2 = 5.5
y = (3 - 9) / 2 = -6/2 = -3

So M (5.5, -3)




2. Calculate the vector AM:

AM = M - A = (5.5, -3) - (-2, -3) = (3.5, 0)

This will be your


3. Find the equation:

I don't know which equation you're used to having to write, but I presume it's the y = mx + b
So m can be calculated dividing the ordinate by the abscess of the vector AM.

m = 0 / 3.5 = 0

m = 0 because A and M have the same ordinate so the line will be parallel to the x-axis, which means all of the points will have the same ordinate as well (x doesn't vary).

y = 0x + b <=> y = b

b is the ordinate of the point that intercepts the y-axis, which is -3 in this case.

You can now conclude that AM can be defined by the equation y = -3





c)
1. Find the gradient of AB:

You already calculated the vector AB before - AB (3, 6)
m = 6 / 3 = 2


1. Find the equation:

Now that you know the m, you know that y = 2x + b. Since the line must contain the point C, all you need to do is replace x and y with the coordinates of C and find b:

y = 2x + b
-9 = 2 (10) + b
-9 = 20 + b
b = -29

So the equation is y = 2x - 29





d)
1. Find the equation of AC, just like in the previous questions:

AC = C - A = (10, -9) - (-2, -3) = (12, -6)

m = -6/12 = -1/2 = -0.5

y = -0.5 x + b
-3 = -0.5 (-2) +b
-3 = 1 + b
b = -4

y = -0.5x - 4




2. Find the coordinates of E:

If E cuts AC at the y-axis, it means its abscissa is 0. You can replace it in the equation and calculate the y. However, if it cuts it in the y-axis, its ordinate is b. Which means E (0, -4).





Please correct me if anything's wrong. I added the graphics since I always find it easier to understand when we can see what we're working with. Anyway, I hope this helped you, and if you didn't understand something or still have questions feel free to ask!

[Rinsea]

@aequilibrium thank you so much for the very complete solution and answer!! I noticed that you used a completely different method to solve the first few questions but it's okay now I understand the lesson (ish)

Can someone else help me? Is there another way to solve this equation?

[aequilibrium]

@SilverIcyCat - You're welcome! This is what we always use right now. I remember we used a different method before learning the vectors for some of the questions, but I don't remember it at all, sorry xs


a)
For this one maybe you had to do something like this:
- Calculate the distance between AB, BC and AC with the formula d(A, B) = √ [ (b1 - a1)^2 + (b2 - a2) ] (square root of the whole thing)
- Use the Pythagoras's theorem to find k

Which would look like this:

d(A, B) = √ [ (1 + 2)^2 + (3 + 3)^2 ] = √(3^2 + 6^2) = √45
d(A, C) = √ [ (10 + 2)^2 + (k + 3)^2 ] = √(12^2 + k^2 + 6k + 9) = √(144 + 9 + k^2 + 6k) = √(153 + k^2 + 6k)
d(B, C) = √ [ (10 - 1)^2 + (k - 3)^2 ] = √(9^2 + k^2 - 6k + 9) = √(81 + 9 + k^2 - 6k) = √(90 + k^2 - 6k)

Because they tell you it's right angled at A, you know AB and AC are the two legs and BC is the hypotenuse, so:

BC^2 = AB^2 + AC^2
[ √(90 + k^2 + 6k) ]^2 = [ √(153 + k^2 + 6k) ]^2 + ( √45 )^2
90 + k^2 + 6k = 153 + k^2 - 6k + 45
+ 6k + 6k = 198 - 90
12k = 108
k = 9

You can see I got something wrong somewhere or there is some rule or whatsoever that I missed / don't remember, but you can use this as a guideline and try to do it on your own.

Note: in the theorem you have to write a line over each segment to indicate it's the measure / length of the segments

[Waki]

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[Waki]

I HAVE A QUESTION!
TYPE OF QUESTION: Social studies-AP Human Geography to be exact;;
YOUR QUESTION:So..(This is going to sound real dumb..;;) I have to memorize the names and location of states in the US and Canada. In other words I have to be able to fill in a blank map...I have always struggled to memorize anything, and I would be forever grateful if someone could tell me if there is an easy way to memorize them;;

there are always state songs you can search up on youtube- like the periodic table song (that's how i memorized my periodic tables haha)

[Rinsea]

@SilverIcyCat - You're welcome! This is what we always use right now. I remember we used a different method before learning the vectors for some of the questions, but I don't remember it at all, sorry xs


a)
For this one maybe you had to do something like this:
- Calculate the distance between AB, BC and AC with the formula d(A, B) = √ [ (b1 - a1)^2 + (b2 - a2) ] (square root of the whole thing)
- Use the Pythagoras's theorem to find k

Which would look like this:

d(A, B) = √ [ (1 + 2)^2 + (3 + 3)^2 ] = √(3^2 + 6^2) = √45
d(A, C) = √ [ (10 + 2)^2 + (k + 3)^2 ] = √(12^2 + k^2 + 6k + 9) = √(144 + 9 + k^2 + 6k) = √(153 + k^2 + 6k)
d(B, C) = √ [ (10 - 1)^2 + (k - 3)^2 ] = √(9^2 + k^2 - 6k + 9) = √(81 + 9 + k^2 - 6k) = √(90 + k^2 - 6k)

Because they tell you it's right angled at A, you know AB and AC are the two legs and BC is the hypotenuse, so:

BC^2 = AB^2 + AC^2
[ √(90 + k^2 + 6k) ]^2 = [ √(153 + k^2 + 6k) ]^2 + ( √45 )^2
90 + k^2 + 6k = 153 + k^2 - 6k + 45
+ 6k + 6k = 198 - 90
12k = 108
k = 9

You can see I got something wrong somewhere or there is some rule or whatsoever that I missed / don't remember, but you can use this as a guideline and try to do it on your own.

Note: in the theorem you have to write a line over each segment to indicate it's the measure / length of the segments

Ah there!! Thanks so much ^o^

[contra.]

I HAVE A QUESTION!
TYPE OF QUESTION: Intro to 2D Art
YOUR QUESTION:Does anyone know a good way to memorize the principles of art and their definitions? I don't have a good method to do this.

[Waki]

I HAVE A QUESTION!
TYPE OF QUESTION: Intro to 2D Art
YOUR QUESTION:Does anyone know a good way to memorize the principles of art and their definitions? I don't have a good method to do this.

You can make examples of each, or create Flashcards (which is just straight up memorizing.) what I do when it comes to straight up memorizing IS writing the definition and rewriting it lots of times. If you ALSO RELATE the definition to the word, and try to understand it instead of just memorizing, it should also be easier

[Waki]

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[Silver Pandorica]

Hope it's okay if I mark this <3