Question 214194
Q.1: Find the lengths of the sides of the triangle with vertices A(-4,0), B(3,4) and C(4,1).show that the triangle ABC is isosceles?

<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5),
locate(-5,0+.5,"A(-4,0)"), locate(3,4.5,"B(3,4)"), locate(4,1,"C(4,1)"),
triangle(-4,0,3,4,4,1) )}}}

There are two ways. It will be instructive to learn
both methods.

Find the length of AB by drawing this blue
right triangle with AB as its hypotenuse: 

{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5),
locate(-5,0+.5,"A(-4,0)"), locate(3,4.5,"B(3,4)"), locate(4,1,"C(4,1)"),
triangle(-4,0,3,4,4,1), blue(line(-4,0,3,0)), blue(line(3,0,3,4)),
blue(line(3,4,-4,0))
 )}}}

The bottom leg of the blue triangle is obviously 7 units long.
(The bottom leg looks purple since it's blue over a red x-axis).
You can just count them. And the right leg of the blue triangle 
is obviously 4 units tall.  You can tell by looking at the units 
of the y axis.

Now AB is the hypotenuse of the blue right triangle. So we use
the Pythagorean theorem:

{{{AB^2=7^2+4^2}}}
{{{AB^2=49+16}}}
{{{AB^2=65}}}
{{{AB=sqrt(65)}}}

Now we erase the blue right triangle and draw a
green one:

We find the length of AC by drawing this green
right triangle with AC as its hypotenuse:

{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5),
locate(-5,0+.5,"A(-4,0)"), locate(3,4.5,"B(3,4)"), locate(4,1,"C(4,1)"),
triangle(-4,0,3,4,4,1), green(line(-4,0,4,0)), green(line(4,0,4,1)),
green(line(4,1,-4,0))
 )}}} 

The bottom leg of the green triangle is obviously 8 units long.
And the right leg of the green triangle is obviously 1 unit tall.

Now AC is the hypotenuse of the green right triangle. So we use
the Pythagorean theorem:

{{{AC^2=8^2+1^2}}}
{{{AC^2=64+1}}}
{{{AC^2=65}}}
{{{AC=sqrt(65)}}}

So the lengths of AB and AC are both equal to{{{sqrt(65)}}}.
Therefore triangle ABC is isosceles.

HERE'S THE SECOND WAY;

Use the distance formula to find AB:

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}
{{{AB=sqrt((3-(-4))^2+(4-0)^2)}}}
{{{AB=sqrt((3+4))^2+(4)^2)}}}
{{{AB=sqrt(7^2+4^2)}}}
{{{AB=sqrt(49+16)}}}
{{{AB=sqrt(65)}}}

Use the distance formula to find AC:

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}
{{{AC=sqrt((4-(-4))^2+(1-0)^2)}}}
{{{AC=sqrt((4+4)^2+(1)^2)}}}
{{{AC=sqrt(8^2+1^2)}}}
{{{AC=sqrt(64+1)}}}
{{{AC=sqrt(65)}}}

So the lengths of AB and AC are both equal to{{{sqrt(65)}}}.
Therefore triangle ABC is isosceles.

Edwin</pre>