+42 Find the distance between the points \((1,1)\) and \((4,7)\) . Express your answer in simplest radical form.
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.
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+208 You can use the distance formula:
\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\:}\)
\(=\sqrt{\left(4-1\right)^2+\left(7-1\right)^2}\)
\(=\sqrt{9+36}\)
\(=\sqrt{45}\)
\(=3\sqrt{5}\)
+208 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, \(AB^2=AC^2+BC^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\)
\(\:AB=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
+118608 Thanks JKP
I know you got the pic from somewhere and it is a good diagram. Give yourself a point 
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.
+118608 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.
