Collinearity Condition

If three points A, B and C are collinear and B lies between A and C, then,

  • AB + BC = AC.  AB, BC, and AC can be calculated using the distance formula.
  • The ratio in which B divides AC, calculated using section formula for both the x and y coordinates separately will be equal.
  • Area of a triangle formed by three collinear points is zero.

NOTES

(I) Four points will form:

(a) A parallelogram if it’s opposite sides is equal, but diagonals are unequal.

(b) A rectangle if opposite sides is equal and two diagonals are also equal.

(c) A rhombus if all the four sides are equal, but diagonals unequal,

(d) A square if all sides are equal and diagonals are also equal.

(II) Three points will form:

(a) An equilateral triangle if all the three sides are equal.

(b) An isosceles triangle if any two sides are equal.

(c) A right angled triangle if sum of square of any two sides is equal to square of the third side.

(d) A triangle if sum of any two sides (distances) is greater than the third side (distance).

(III) Three points A, B and C are collinear or lie on a line if one of the following holds

(i) AB + BC — AC

(ii) AC + CB AB

(iii) CA + AB CB.

Example
Find the distance of the point (-3, 4) from the x-axis.
Solution:
B(-3, 0), A (-3, 4)
\begin{array}{l}AB = \sqrt {{{( - 3 + 3)}^2} + {{(4 - 0)}^2}} \\AB = \sqrt {{{(4)}^2} = 4} \end{array}

Example.
If the points A(x, 2), B(-3, 4) and C(7, -5) are collinear, then find the value of x.
Solution:
When the points are collinear,

\begin{array}{l}{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0\\x( - 4 - ( - 5) + ( - 3)( - 5 - 2) + 7(2 - ( - 4)) = 0\\x(1) + 21 + 42 = 0\\x + 63 = 0\\\therefore x =  - 63\end{array}

Example
In which quadrant the point P that divides the line segment joining the points A (2, -5) and B(5,2) in the ratio 2 : 3 lies?
Solution:

P = \left( {\dfrac{{2(5) + 3(2)}}{{2 + 3}},\dfrac{{2(2) + 3( - 5)}}{{2 + 3}}} \right)

P = \left( {\dfrac{{16}}{5},\dfrac{{ - 11}}{5}} \right) IV Quadrant

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