Distance of Coordinates

**want2math** · January 4th 2010, 04:32 AM

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?

**dedust** · January 4th 2010, 04:51 AM

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 $(x_{1},y_{1})$ and $(x_{2},y_{2})$ : $d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$

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

try number 2

**want2math** · January 4th 2010, 05:09 AM

Originally Posted by dedust

the distance for points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ : $d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$

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

try number 2

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

What am I doing wrong?

**dedust** · January 4th 2010, 05:12 AM

Originally Posted by want2math

$\sqrt{(p-3q-p+4q)^2 + (p-p-q)^2}$
$\sqrt{(p-p+4q-3q)^2 + (p-p-q)^2}$ $\leftarrow$ should be $\sqrt{(p-p-4q-3q)^2 + (p-p+q)^2}$

What am I doing wrong?

. . .

**HallsofIvy** · January 4th 2010, 05:17 AM

Originally Posted by want2math

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

What am I doing wrong?

(p+4q)- (p- 3q)= (p-p)+ (4q+ 3q)= 7q
(p- q)- (p)= -q but (-q)^2= q^2.

**e^(i*pi)** · January 4th 2010, 05:17 AM

Originally Posted by want2math

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

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. $a-(a-b) = a-a+b = b$ . Remember subtracting a negative is like adding a positive

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

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

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

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

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

**want2math** · January 4th 2010, 05:21 AM

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. $a-(a-b) = a-a+b = b$ . Remember subtracting a negative is like adding a positive

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

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

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

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

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

Thanks for the amazing post.

This post clears any doubt in my mind.

Again, thank you all.

Thread: Distance of Coordinates

LinkBack

Thread Tools

Display

Distance of Coordinates

Similar Math Help Forum Discussions

Transforming integral in cartesian coordinates to polar coordinates

Distance between Coordinates

Algorithm to find integer coordinates in 2-d plane at a distance 5 from each other

Distance b. Points (Polar Coordinates)

convert polar coordinates to rectangular coordinates

Search Tags