HOW TO SHOW THAT THE GIVEN POINTS ARE COLLINEAR USING SECTION FORMULA

Question 1 :

Using section formula, show that the points A(7, −5), B(9, −3) and C(13, 1) are collinear.

Solution :

Let point C(13, 1) divide the other ratio k : 1. So we can take as m = k and n = 1 

So x₁ = 7, y₁ = - 5, x₂ = 9, y₂ = - 3

We know that section formula is given by 

(mx₂ + nx₁ /m + n , my₂ + ny₁ /m + n)

(13 , 1) = (9k + 7(1) / k +1 , -3k + (-5)1 / k + 1)

Equating the co ordinates to the respective numbers we get

9k + 7 / k + 1 = 13 ,  -3k - 5 / k + 1 = 1

  - 4k = 6                      -4k = 6

   k = - 3/2                   k = - 3/2

Since k is negative , it divides externally in the ratio 3 :2. Thus the points A,B and C are collinear.

Question 2 :

A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (−2, −3) and (2, 1) respectively, then find the coordinates of C.

Solution :

Let the length of line segment AB is x units.

It is given that AB is increased by 25% extended upto C.

Hence, BC  =  (25/100) x  =  x/4

AB is externally divided into m : n i.e 5 : 1

By Section formula for external division

=  (mx₂ - nx₁ /m - n , my₂ - ny₁ /m - n)

=  (5(2) - 1(-2))/(5-1), (5(1) - 1(-3))/(5-1)

=  (10 + 2)/4, (5 + 3)/4

=  12/4, 8/4

=  (3, 2)

Hence the required point C is (3, 2).

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