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3 (i) Verify the following:
(0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

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Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(1-0)^2+(6-7)^2+(-6-(-10))^2}$

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

$BC=\sqrt{(4-1)^2+(9-6)^2+(-6-(-6))^2}$

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

$CA=\sqrt{(4-0)^2+(9-7)^2+(-6-(-10))^2}$

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see $AB=BC\neq CA$

Hence we can say that ABC is an isosceles triangle.

Posted by

Pankaj Sanodiya

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