Question

Find the coordinates of the points of tri section of the line segment joining the points (-3,-5) and (-6,-8)​

Answer 1

Answer :-

(-4,-6) and (-5,-7)

Solution :-

Let the points of line segment be P(-3,-5) and Q(-6,-8)

Let the points of trisection be A, B

A divides the line segment internally in the ratio of 1 : 2

Using section formula

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

P(-3,-5) Q(-6,-8) m : n = 1 : 2

Here,

• m = 1
• n = 2
• x2 = - 6
• x1 = - 3
• y2 = - 8
• y1 = - 5

Substituting the values

$\implies A(x , y) = \bigg( \dfrac{1( - 6) + 2( - 3)}{1 + 2} , \dfrac{1( - 8)+ 2( - 5) }{1 + 2} \bigg)$

$\implies A(x , y) = \bigg( \dfrac{ - 6 - 6}{3} , \dfrac{ - 8 - 10 }{3} \bigg)$

$\implies A(x , y) = \bigg( \dfrac{ - 12}{3} , \dfrac{ - 18 }{3} \bigg)$

$\implies A(x , y) = ( - 4 , - 6 )$

B divides the line segment internally in the ratio of 2 : 1

Using section formula

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

P(-3,-5) Q(-6,-8) m : n = 2 : 1

Here,

• m = 2
• n = 1
• x2 = - 6
• x1 = - 3
• y2 = - 8
• y1 = - 5

Substituting the values

$\implies B(x , y) = \bigg( \dfrac{2( - 6) + 1( - 3)}{2 + 1} , \dfrac{2( - 8)+ 1( - 5) }{2 + 1} \bigg)$

$\implies B(x , y) = \bigg( \dfrac{ - 12 - 3}{3} , \dfrac{ - 16 - 5 }{3} \bigg)$

$\implies B(x , y) = \bigg( \dfrac{ - 15}{3} , \dfrac{ - 21 }{3} \bigg)$

$\implies B(x , y) = ( - 5 , - 7)$

Therefore the coordinates of trisection are (- 4, - 6) and (-5, - 7)