+0  
 
0
167
9
avatar+42 

Find the distance between the points \((1,1)\) and \((4,7)\) . Express your answer in simplest radical form.

 Jun 6, 2021
 #1
avatar
+2

 

Draw it.  You see that you can form a right triangle with the line between the two points is the hypotenuse. 

 

The bottom leg is         4 – 1 = 3 units.  

The right hand leg is    7 – 1 = 6 units.  

 

Now all you have to do is use Pythagoras' Theorem.  

 

32 + 62 = x2      

 

so x2 = 45 and x (which is the hypotenuse)  =  3 • sqrt(5)   This is the distance between those two points.  

.

 Jun 6, 2021
 #2
avatar+208 
+1

You can use the distance formula:

 \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\:}\)

\(=\sqrt{\left(4-1\right)^2+\left(7-1\right)^2}\)

\(=\sqrt{9+36}\)

\(=\sqrt{45}\)

\(=3\sqrt{5}\)

 Jun 6, 2021
 #3
avatar+115441 
0

It helps some people enormously when they understand that the distance formula IS JUST Pythagoras's theorem.

 

The distance in the y direction, y2-y1  IS  the vertial rise.  And x2-x1 IS the horizonal distance

 

If anyone wants me to explain better I can. Just ask.

Melody  Jun 6, 2021
 #7
avatar+42 
+1

Thank you so much for taking time out of your day it help me with this question.

Scythgirl  Jun 7, 2021
 #4
avatar+208 
+2

Here's the explanation:

 

 

I couldn't find a picture, but the right angle corner is point C.  

So, C's coordinates are \(\left(x_2,\:y_1\right)\)

BC is \(\left(y_2-y_1\right)\)

AC is \(\left(x_2-x_1\right)\)

so, \(AB^2=AC^2+BC^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\)

\(\:AB=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

 Jun 6, 2021
 #5
avatar+115441 
+1

Thanks JKP

 

I know you got the pic from somewhere and it is a good diagram.   Give yourself a point   wink

 

 

Later on, when you are able to draw such a pic yourself it would be better if there were actual number points on the diagram.

It makes it much easier for kids to understand what is going on.

Melody  Jun 7, 2021
edited by Melody  Jun 7, 2021
edited by Melody  Jun 7, 2021
 #6
avatar+115441 
+1

I've just seen some of your other answers, I think I have underestimated your abilities.  Sorry.

Maybe you can already produce a pic like this yourself.

If you do not already have the skill, try playing with GeoGebra. 

Geogebra is a free program, it is what I use for most of my illustrative pics.

It took me a while to learn, and I am still learning, (just trial and error) but it is really fun to use.

Melody  Jun 7, 2021
 #8
avatar+208 
+1

Do you mean graphing with desmos?

JKP1234567890  Jun 8, 2021
 #9
avatar+115441 
0

No

I mean with GeoGebra

Melody  Jun 9, 2021

33 Online Users

avatar
avatar
avatar