Question

Show that the points $P( - 2,3,5),Q(1,2,3){\text{ and R}}(7,0, - 1)$ are collinear.

Answer 1

Hint- In this question we have coordinates of three points, the points must be called as collinear points if and only if all the points lie on a single line and by distance formula we will get the distance between two points which is given as: $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $. By using these two properties we will get the answer.

Complete step-by-step answer:
We will use the following figure to solve the problem.

Here we are using distance formula
$D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $
Point $P,Q{\text{ and }}R$ are collinear if they lie on a line.
The distance between points P and Q is
$
  PQ = \sqrt {{{(1 + 2)}^2} + {{(2 - 3)}^2} + {{(3 - 5)}^2}} \\
   = \sqrt {{3^2} + {{( - 1)}^2} + {{( - 2)}^2}} \\
   = \sqrt {9 + 1 + 4} \\
   = \sqrt {14} \\
 $
The distance between points Q and R is
$
  QR = \sqrt {{{(7 - 1)}^2} + {{(0 - 2)}^2} + {{( - 1 - 3)}^2}} \\
   = \sqrt {{6^2} + {{( - 2)}^2} + {{( - 4)}^2}} \\
   = \sqrt {36 + 4 + 16} \\
   = 2\sqrt {14} \\
 $
The distance between points P and R is
$
  PR = \sqrt {{{(7 + 2)}^2} + {{(0 - 3)}^2} + {{( - 1 - 5)}^2}} \\
   = \sqrt {{9^2} + {{( - 3)}^2} + {{( - 6)}^2}} \\
   = \sqrt {81 + 9 + 36} \\
   = 3\sqrt {14} \\
 $
Hence,
$
  PQ + QR = RP \\
   = \sqrt {14} + 2\sqrt {14} = 3\sqrt {14} \\
 $
Therefore the points are collinear.

Note- In this type of questions remember the distance formula and the basic concept that if lines are points are collinear they lie on the same line and the sum of distance of two lines is equal to the distance of the third line. This problem can also be solved by using the formula to calculate the area of the triangle. As we know, if the points are collinear then the area of the triangle is zero.