Question

Show that the point (2,0),(-2,0)and (0,2)are the vertices of a triangle also state with reason the type of the triangle

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Given,

Location of 1 st point - A(2,0)

Location of 2nd point - B(-2,0)

Location of 3rd point - A(0,2)

The distance between two points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ is given by:

$d(P,Q) =\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Therefore distance between A and B

$d(A,B) =\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\\\=\sqrt{(-2-2)^{2}+(0-0)^{2}}\\\\=\sqrt{(-4)^{2}}\\\\=\sqrt{16}\\\\=4$

$d(A,C) =\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\\\=\sqrt{(0-2)^{2}+(2-0)^{2}}\\\\=\sqrt{(-2)^{2}+(2)^{2}}\\\\=\sqrt{4+4}\\\\=\sqrt{8}$

$d(B,C) =\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\\\=\sqrt{(0-(-2))^{2}+(2-0)^{2}}\\\\=\sqrt{(2)^{2}+(2)^{2}}\\\\=\sqrt{4+4}\\\\=\sqrt{8}$

Triangle formed by this three point are Isosceles triangle. Because two sides are equal. $AC=BC=\sqrt{8}$