Question

Show that the points $(2,3, - 4),(1, - 2,3)$ and $(3,8, - 11)$ are collinear.

Answer 1

We use the concept of collinear points and that three points in general form a triangle and if the points are collinear then the area of the triangle formed by collinear points is equal to zero. Use Heron’s formula to calculate the area of the triangle formed.
Distance point formula for two points $\left( {{x_1},{y_1},{z_1}} \right);\left( {{x_2},{y_2},{z_2}} \right)$is given by the formula $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

• Three points are said to be collinear if they lie on the same line.
• Heron’s formula gives us the formula for the area of a triangle whose three side lengths are given. Given $a,b,c$ as three sides of triangle, we find the semi-perimeter first $s = \dfrac{1}{2}(a + b + c)$ and then
  Area of triangle $= \sqrt {s(s - a)(s - b)(s - c)}$

Step-By-Step answer:
We are given three points $(2,3, - 4),(1, - 2,3)$ and $(3,8, - 11)$
Let points be named as $A(2,3, - 4),B(1, - 2,3)$ and $C(3,8, - 11)$
We calculate the distance or length of sides using distance point formula
Side AB:
AB is made by joining coordinates $A(2,3, - 4),B(1, - 2,3)$
$\Rightarrow AB = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( { - 2 - 3} \right)}^2} + {{\left( {3 - ( - 4)} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow AB = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( {3 + 4} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( 7 \right)}^2}}$
Square the numbers under the root
$\Rightarrow AB = \sqrt {1 + 25 + 49}$
$\Rightarrow AB = \sqrt {75}$
$\Rightarrow AB = \sqrt {25 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow AB = \sqrt {{5^2} \times 3}$
Cancel square root by square power
$\Rightarrow AB = 5\sqrt 3$ … (1)
Side BC:
BC is made by joining coordinates $B(1, - 2,3),C(3,8, - 11)$
$\Rightarrow BC = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {8 - ( - 2)} \right)}^2} + {{\left( { - 11 - 3} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow BC = \sqrt {{{\left( 2 \right)}^2} + {{\left( {8 + 2} \right)}^2} + {{\left( { - 14} \right)}^2}}$
$\Rightarrow BC = \sqrt {{{\left( 2 \right)}^2} + {{\left( {10} \right)}^2} + {{\left( { - 14} \right)}^2}}$
Square the numbers under the root
$\Rightarrow BC = \sqrt {4 + 100 + 196}$
$\Rightarrow BC = \sqrt {300}$
$\Rightarrow BC = \sqrt {100 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow BC = \sqrt {{{10}^2} \times 3}$
Cancel square root by square power
$\Rightarrow BC = 10\sqrt 3$ … (2)
Side CA:
AB is made by joining coordinates $C(3,8, - 11),A(2,3, - 4)$
$\Rightarrow CA = \sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( {3 - 8} \right)}^2} + {{\left( { - 4 - ( - 11)} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow CA = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( { - 4 + 11} \right)}^2}}$
$\Rightarrow CA = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( 7 \right)}^2}}$
Square the numbers under the root
$\Rightarrow CA = \sqrt {1 + 25 + 49}$
$\Rightarrow CA = \sqrt {75}$
$\Rightarrow CA = \sqrt {25 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow CA = \sqrt {{5^2} \times 3}$
Cancel square root by square power
$\Rightarrow CA = 5\sqrt 3$ … (3)
Length of sides of a triangle is $5\sqrt 3 ;10\sqrt 3 ;5\sqrt 3$
Let $a = 5\sqrt 3 ,b = 10\sqrt 3 ,c = 5\sqrt 3$
We find the semi-perimeter of the triangle;
$\Rightarrow s = \dfrac{1}{2}(a + b + c)$
Substitute the values of ‘a’, ‘b’ and ‘c’
$\Rightarrow s = \dfrac{1}{2}(5\sqrt 3 + 10\sqrt 3 + 5\sqrt 3 )$
Take $\sqrt 3$common and add the terms
$\Rightarrow s = \dfrac{1}{2}(\sqrt 3 (5 + 10 + 5))$
$\Rightarrow s = \dfrac{1}{2} \times 20\sqrt 3$
Cancel same factors from numerator and denominator
$\Rightarrow s = 10\sqrt 3$ … (4)
Then using Herons formula to find the area of a triangle
$\Rightarrow$ Area of triangle $= \sqrt {s(s - a)(s - b)(s - c)}$
Substitute the value of ‘a’, ‘b’, ’c’ and ‘S’ in the formula
$\Rightarrow$ Area of triangle $= \sqrt {10\sqrt 3 (10\sqrt 3 - 5\sqrt 3 )(10\sqrt 3 - 10\sqrt 3 )(10\sqrt 3 - 5\sqrt 3 )}$
$\Rightarrow$ Area of triangle $= \sqrt {10\sqrt 3 (5\sqrt 3 )(0)(5\sqrt 3 )}$
$\Rightarrow$ Area of triangle $= \sqrt 0$
$\Rightarrow$ Area of triangle $= 0$
$\therefore$ Area of triangle formed by three points is equal to 0.

$\therefore$ Points $(2,3, - 4),(1, - 2,3)$ and $(3,8, - 11)$ are collinear.

Note: Students many times try to draw an altitude and then find the height of the triangle using Pythagoras theorem and then use the formula for area, but this is very complex and a long process so students should avoid this. When writing the value of sides, always simplify the square root i.e. take out as many terms possible from the root.