Question

Find coordinates of a point which divides the line segment (6,-10) and (3, -8) internally in a ratios 2:7​

Answer 1

Topic :-

Co-ordinate system

Given :-

The point (6 , - 10) and (3 , -8 ) divides the line segment in ratio 2 : 7

To find :-

• The co-ordinates of a point

Solution:-

If the point (x , y ) divides the line segment ${(x_1, y_1)}$ and ${(x_2,y_2)}$ in ratio m:n then

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

So,

${(x_1,y_1) = (6 , -10)}$

${(x_2, y_2) =(3 , -8)}$

m : n = 2:7

Substuiting the values

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

${(x,y)} = \bigg(\dfrac{2(3)+7(6)}{2+7} , \dfrac{2(-8)+7(-10)}{2+7}\bigg)$

${(x,y)} = \bigg(\dfrac{6+42}{9} , \dfrac{-16-70}{9}\bigg)$

${(x,y)} = \bigg(\dfrac{48}{9} , \dfrac{-86}{9}\bigg)$

So, the point (48/9, -86/9) divides the line segemnt (6,-10) and (3, -8) internally in a ratios 2:7

Know more :-

Distance formula:-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Centroid formula:-

$\bigg(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\bigg)$

Section formula External division

$\bigg(\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}\bigg)$

Mid point formula:-

$\bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2}\bigg)$