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Home » Show that the points A(1, -1, -5), B(3, 1,3) and C(9, 1, -3) are the vertices of an equilateral triangle.
Show that the points A(1, -1, -5), B(3, 1,3) and C(9, 1, -3) are the vertices of an equilateral triangle.
CBSE | Class 11 | Exercise 26B | Maths | RS Aggarwal | Three Dimensional Geometry
Show that the points A(1, -1, -5), B(3, 1,3) and C(9, 1, -3) are the vertices of an equilateral triangle.

Answer:

(x1,y1,z1) = (1, -1, -5)

(x2,y2,z2) = (3, 1,3)

(x3,y3,z3) = (9, 1, -3)

\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\ => \sqrt {{{( 3 - 1)}^2} + {{(1 - (-1))}^2} + {{( 3 - (-5))}^2}} \\ => \sqrt {4 + 4 + 64} \\ => \sqrt {72} => 6\sqrt {2} \end{array}

\begin{array}{l} Length BC = \sqrt {{{({x_3} - {x_2})}^2} + {{({y_3} - {y_2})}^2} + {{({z_3} - {z_2})}^2}} \\ => \sqrt {{{( 9 - 3)}^2} + {{(1 - 1))}^2} + {{( -3 - 3))}^2}} \\ => \sqrt {36 + 0 + 36} \\ => \sqrt {72} => 6\sqrt {2} \end{array}

\begin{array}{l} Length AC = \sqrt {{{({x_3} - {x_1})}^2} + {{({y_3} - {y_1})}^2} + {{({z_3} - {z_1})}^2}} \\ => \sqrt {{{( 9 - 1)}^2} + {{(1 - (-1))}^2} + {{( -3 - (-5))}^2}} \\ => \sqrt {64 + 4 + 4} \\ => \sqrt {72} => 6\sqrt {2} \end{array}

AB = BC = AC

Hence, the points A, B, C make an equilateral triangle.

 

 

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