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Verify that $(0,7,10),\:(-1,6,6)\:and\:(-4,9,6)$ are vertices of a right angled 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 $\overline {AC}=\sqrt {(-4-0)^2+(9-7)^2+(6-10)^2}=\sqrt{16+4+16}=\sqrt {36}=6$
$\therefore$ From Pythogorous throrem we can say that $ABC$ is rt. angled $\Delta$
answered Apr 17, 2014 by rvidyagovindarajan_1

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