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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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