Question

8. Points A, B, C and D divide the line segment join-
ing the point (5,-10) and the origin in five equal parts.
Find the co-ordinates of A, B, C and D.

Answer 1

Solution :-

A line segment is formed by joining origin and a point (5,-10)

The coordinates of origin are (0,0)

A, B, C, D divide the line segment into 5 equal parts

A divides the line segment internally in the ratio of 1 : 4

(0,0) (5,-10) m : n = 1 : 4

By using section formula :

$P(x,y) = \bigg( \dfrac{mx_2 + nx_1 }{m + n} , \dfrac{my_2 + ny_1 }{m + n} \bigg)$

Here, x₁ = 0, x₂ = 5, y₁ = 0 y₂ = - 10, m = 1, n =

$\implies A(x,y) = \bigg( \dfrac{1(5)+ 4(0) }{1 + 4} , \dfrac{1( - 10) +4(0) }{1 + 4} \bigg)$

$\implies A(x,y) = \bigg( \dfrac{5 }{5} , \dfrac{ - 10}{5} \bigg) = (1, - 2)$

B divides the line segment internally in the ratio of 2 : 3

(0,0) (5, - 10) m : n = 2 : 3

Here, x₁ = 0, x₂ = 5, y₁ = 0 y₂ = - 10, m = 2, n = 3

$\implies B(x,y) = \bigg( \dfrac{2(5)+ 3(0) }{2 + 3} , \dfrac{2( - 10) +3(0) }{2 + 3} \bigg)$

$\implies B(x,y) = \bigg( \dfrac{10 }{5} , \dfrac{ - 20}{5} \bigg) = (2 , - 4)$

C divivides line segment internally in the ratio of 3 : 2

(0,0) (5,-10) m : n = 3 : 2

Here, x₁ = 0, x₂ = 5, y₁ = 0 y₂ = - 10, m = 3, n = 2

$\implies C(x,y) = \bigg( \dfrac{3(5)+ 2(0) }{3 + 2} , \dfrac{3( - 10) +2(0) }{3 + 2} \bigg)$

$\implies C(x,y) = \bigg( \dfrac{15 }{5} , \dfrac{ - 30 }{5} \bigg) = (3, - 6)$

D divides line segment internally in the ratio of 4 : 1

(0,0) (5,-10) m : n = 4 : 1

Here, x₁ = 0, x₂ = 5, y₁ = 0 y₂ = - 10, m = 4, n = 1

$\implies D(x,y) = \bigg( \dfrac{4(5)+ 1(0) }{4 + 1} , \dfrac{4( - 10) +1(0) }{4 + 1} \bigg)$

$\implies D(x,y) = \bigg( \dfrac{20 }{5} , \dfrac{ - 40}{5} \bigg) = (4 , - 8)$

Therefore the coordinates of A, B, C, D are (1,-2), (2, - 4), (3, - 6), (4, - 8) respectively.