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Verify that $(0,7,-10),\:(1,6,-6)\:and\:(4,9,-6)$ are vertices of an isosceles triangle.

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  • distance between the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Let $A(0,7,-10),\:B(1,6,-6)\:and\:C(4,9,-6)$ be the given three points.
We know that the distance between the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by
$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Distance $\overline {AB}=\sqrt {(1-0)^2+(6-7)^2+(-6+10)^2}=\sqrt{1+1+16}=\sqrt {18}=3\sqrt 2$
Distance $\overline {BC}=\sqrt {(4-1)^2+(9-6)^2+(-6+6)^2}=\sqrt{9+9+0}=\sqrt {18}=3\sqrt {2}$
Distance $\overline {AC}=\sqrt {(4-0)^2+(9-7)^2+(-6+10)^2}=\sqrt{16+4+16}=\sqrt {36}=6$
$\Rightarrow\:\overline {AB}=\overline {BC}$ and sum of any two sides is $\neq$ third side
$\therefore$ The three points are not collinear and they form an isosceles triangle.
answered Apr 17, 2014 by rvidyagovindarajan_1
edited Apr 17, 2014 by rvidyagovindarajan_1
 

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