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2. Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.

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Given three points, A = (-2, 3, 5), B = (1, 2, 3) and C = (7, 0, -1)

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-(-2))^2+(2-3)^2+(3-5)^2}

AB=\sqrt{(3)^2+(-1)^2+(-2)^2}

AB=\sqrt{9+1+4}

AB=\sqrt{14}

The distance BC:

BC=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}

BC=\sqrt{36+4+16}

BC=\sqrt{56}

BC=2\sqrt{14}

The distance CA

CA=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}

CA=\sqrt{81+9+36}

CA=\sqrt{126}

CA=3\sqrt{14}

As we can see here, 

AB+BC=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=AC

Hence, we can say that points A,B and C are colinear.

Posted by

Pankaj Sanodiya

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