1. ## Distance of Coordinates

I'm really weak at maths and I have to study for my upcoming exams. I can't seem to find the distance of these coordinates.

1. (p,q) and (q,p)
2. (p+4q, p-q) and (p-3q, p)

I think in the first one I have to open the brackets using the a2-2ab+b2 formula and in the second question, I am doing it normally but my answer stops at q2+q2 under square root. Help?

2. Originally Posted by want2math
I'm really weak at maths and I have to study for my upcoming exams. I can't seem to find the distance of these coordinates.

1. (p,q) and (q,p)
2. (p+4q, p-q) and (p-3q, p)

I think in the first one I have to open the brackets using the a2-2ab+b2 formula and in the second question, I am doing it normally but my answer stops at q2+q2 under square root. Help?
the distance for points $\displaystyle (x_{1},y_{1})$ and $\displaystyle (x_{2},y_{2})$ : $\displaystyle d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$

1. $\displaystyle d = \sqrt{(q - p)^2 + (p - q)^2} = \sqrt{(q - p)^2 + (q - p)^2} = \sqrt{2(q - p)^2} = (q - p) \sqrt{2}$

try number 2

3. Originally Posted by dedust
the distance for points $\displaystyle (x_{1},y_{1})$ and $\displaystyle (x_{2},y_{2})$ : $\displaystyle d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$

1. $\displaystyle d = \sqrt{(q - p)^2 + (p - q)^2} = \sqrt{(q - p)^2 + (q - p)^2} = \sqrt{2(q - p)^2} = (q - p) \sqrt{2}$

try number 2
$\displaystyle \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{q^2 + q^2}$

What am I doing wrong?

4. Originally Posted by want2math
$\displaystyle \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}$ $\displaystyle \leftarrow$ should be $\displaystyle \sqrt{(p-p-4q-3q)^2 + (p-p+q)^2}$

What am I doing wrong?
. . .

5. Originally Posted by want2math
$\displaystyle \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{q^2 + q^2}$

What am I doing wrong?
(p+4q)- (p- 3q)= (p-p)+ (4q+ 3q)= 7q
(p- q)- (p)= -q but (-q)^2= q^2.

6. Originally Posted by want2math
$\displaystyle \sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}$
$\displaystyle \sqrt{q^2 + q^2}$

What am I doing wrong?
It's easier to follow if you use brackets to show which term is which. I can't follow that above so I'll detail my way below

It seems you're getting confused with negative signs. $\displaystyle a-(a-b) = a-a+b = b$. Remember subtracting a negative is like adding a positive

$\displaystyle \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2}$

$\displaystyle (p-3q)-(p+4q) = p-3p-p-4q = -7q$

$\displaystyle p-(p-q) = p-p+q = q$

$\displaystyle \therefore \: \: \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2} = \sqrt{(-7q)^2 + q^2} = \sqrt{50q^2} = \sqrt{2 \cdot 25 \cdot q^2}$

25 and q^2 are both squares so we can remove them from the surd: $\displaystyle 5q \sqrt{2}$

7. Originally Posted by e^(i*pi)
It's easier to follow if you use brackets to show which term is which. I can't follow that above so I'll detail my way below

It seems you're getting confused with negative signs. $\displaystyle a-(a-b) = a-a+b = b$. Remember subtracting a negative is like adding a positive

$\displaystyle \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2}$

$\displaystyle (p-3q)-(p+4q) = p-3p-p-4q = -7q$

$\displaystyle p-(p-q) = p-p+q = q$

$\displaystyle \therefore \: \: \sqrt{[(p-3q)-(p+4q)]^2 + [p-(p-q)]^2} = \sqrt{(-7q)^2 + q^2} = \sqrt{50q^2} = \sqrt{2 \cdot 25 \cdot q^2}$

25 and q^2 are both squares so we can remove them from the surd: $\displaystyle 5q \sqrt{2}$
Thanks for the amazing post.

This post clears any doubt in my mind.

Again, thank you all.