A(1,3)
B(2,6)

those are the points on the graph and they want me to fine the distancebetween points A and B... welll i like worked out the problem step by step and i keep getting i differnet answer that is in the book!!! i need som serious help! could you please!!!

i keep getting 10 for an answer and the answer says it is 30. and i can not see where i keep making the mistake can you please help me. THANK YOU SOO MUCH!

Minthie Jo

2. Originally Posted by mj007

A(1,3)
B(2,6)

those are the points on the graph and they want me to fine the distancebetween points A and B... welll i like worked out the problem step by step and i keep getting i differnet answer that is in the book!!! i need som serious help! could you please!!!

i keep getting 10 for an answer and the answer says it is 30. and i can not see where i keep making the mistake can you please help me. THANK YOU SOO MUCH!

Minthie Jo
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
So:
$d=\sqrt{(2-1)^2+(6-3)^2}$
$d=\sqrt{(1)^2+(3)^2}$
$d=\sqrt{1+9}=\sqrt{10}$.

So I disagree with both you and your book. And your little dog, too! (Ahem. It's late and the Wizard of Oz reference just seemed too good to pass up. Okay okay! I'm going to bed now. )

-Dan

3. Originally Posted by topsquark
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
So:
$d=\sqrt{(2-1)^2+(6-3)^2}$
$d=\sqrt{(1)^2+(3)^2}$
$d=\sqrt{1+9}=\sqrt{10}$.

So I disagree with both you and your book. And your little dog, too! (Ahem. It's late and the Wizard of Oz reference just seemed too good to pass up. Okay okay! I'm going to bed now. )

-Dan

wait a minute how do you disagree with me?? i got 10 too..

4. A(1,3)
B(2,6)

those are the points on the graph and they want me to fine the distancebetween points A and B... welll i like worked out the problem step by step and i keep getting i differnet answer that is in the book!!! i need som serious help! could you please!!!

===
Distance is the hypotenuse between the points. It really helps to draw a sketch to help see this. Create a right triangle with one side being the segment between the x values and the other segment being between the y values of the points. Find the lengths of those 'legs' and Pythagoras rides to the rescue.

Difference in x's = 1 and diff in y's = 3

a^2 + b^2 = c^2
1^2 + 3^2 = c^2
1+9 = c^2 Take SQRT of each side of the eq.

A(1,3)
B(2,6)

those are the points on the graph and they want me to fine the distancebetween points A and B... welll i like worked out the problem step by step and i keep getting i differnet answer that is in the book!!! i need som serious help! could you please!!!

===
Distance is the hypotenuse between the points. It really helps to draw a sketch to help see this. Create a right triangle with one side being the segment between the x values and the other segment being between the y values of the points. Find the lengths of those 'legs' and Pythagoras rides to the rescue.

Difference in x's = 1 and diff in y's = 3

a^2 + b^2 = c^2
1^2 + 3^2 = c^2
1+9 = c^2 Take SQRT of each side of the eq.
Just to make sure the point is driven home, I got $\sqrt{10}$, you said you got 10. Perhaps this is what you meant, but it is not what you said. There is a big difference between the two.

-Dan