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

Scythgirl Jun 6, 2021

#1**+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.

3^{2} + 6^{2} = x^{2}

so x^{2} = 45 and x (which is the hypotenuse) = **3 • sqrt(5)** This is the distance between those two points.

_{.}

Guest Jun 6, 2021

#2**+1 **

You can use the distance formula:

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

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

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

\(=\sqrt{45}\)

\(=3\sqrt{5}\)

JKP1234567890 Jun 6, 2021

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

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

JKP1234567890 Jun 6, 2021

#5**+1 **

**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.

Melody
Jun 7, 2021

#6**+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