Answer:
Triangle ABC has vertices A(0, 0), B(5, 2) and C(7,-3).
Show that ABC is isosceles.
If a triangle has vertices [tex]A(-2,4), \ B(6,2)[/tex] and [tex]C(1,-1)[/tex] then [tex]\triangle ABC[/tex] is an isosceles right triangle.
An Isosceles Right Triangle is a right triangle that have two equal sides.
To know whether it is Isosceles Right Triangle or not we will use ,
Distance formula [tex]= \sqrt{(x_{2} -x_{1})^{2} + (y_{2} -y_{1})^{2}}[/tex]
We have,
Vertices [tex]A(-2,4), \ B(6,2)[/tex] and [tex]C(1,-1)[/tex] of [tex]\triangle ABC[/tex].
So
To find [tex]AB[/tex] coordinates of [tex]A[/tex] and [tex]B[/tex] will be used and so on;
[tex]AB=\sqrt{(6-(-2))^{2} +(2-4)^2} \ =\sqrt{(8)^2+ (-2)^2} \ = \sqrt{64+4} = \sqrt{68}[/tex]
[tex]BC=\sqrt{(1-6)^{2} +(-1-2)^2} \ =\sqrt{(-5)^2+ (-3)^2} \ = \sqrt{25+9} = \sqrt{34}[/tex]
[tex]AC=\sqrt{(1-(-2))^{2} +(-1-4)^2} \ =\sqrt{(3)^2+ (-5)^2} \ = \sqrt{9+25} = \sqrt{34}[/tex]
Here,
[tex]BC=AC[/tex] which means that two sides of [tex]\triangle ABC[/tex] are equal i.e. it is isosceles triangle.
And,
Using Pythagoras theorem;
[tex]AC^2+BC^2=AB^2[/tex]
[tex](\sqrt{34} )^2+(\sqrt{34} )^2=(\sqrt{68} )^2[/tex]
[tex]34+34=68[/tex]
[tex]68=68[/tex]
So, this is satisfying the Pythagoras theorem.
So, this is Isosceles Right Triangle because it is satisfying the Pythagoras theorem and property of isosceles triangle..
Hence, we can say that if a triangle has vertices [tex]A(-2,4), \ B(6,2)[/tex] and [tex]C(1,-1)[/tex] then [tex]\triangle ABC[/tex] is an isosceles right triangle.
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