Question

Q34. Find the co-ordinates of the points of trisection of the line segment joining the point (3, –1) and (6, 8).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given a line segment joining the points (3, - 1) and (6, 8).

Let assume that the given line segment be AB so that

Coordinates of A be ( 3, - 1 )

and

Coordinates of B be ( 6, 8 ).

Let assume that P and Q be the required points which trisect the line segment AB.

So, that, AP = PQ = QB

It means,

P divides AB in the ratio 1 : 2

and

Q divides AB in the ratio 2 : 1.

Case :- 1 When P divides AB in the ratio 1 : 2.

We know,

Section Formula is used to find the coordinates of the point C ( x, y ) which divides the line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) internally in the ratio m : n, then coordinates of C is

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

So, here

$\rm :\longmapsto\:x_1 = 3$

$\rm :\longmapsto\:y_1 = - \: 1$

$\rm :\longmapsto\:x_2= 6 \:$

$\rm :\longmapsto\:y_2= 8 \:$

$\rm :\longmapsto\:m = 1$

$\rm :\longmapsto\:n = 2$

On substituting the values,

$\rm :\longmapsto\:Coordinates \: of \: P = \bigg(\dfrac{1 \times 6 + 2 \times 3}{1 + 2} ,\dfrac{1 \times 8 + 2 \times ( - 1)}{2 + 1}\bigg)$

$\rm :\longmapsto\:Coordinates \: of \: P = \bigg(\dfrac{6 + 6}{3} ,\dfrac{8 - 2}{3} \bigg)$

$\rm :\longmapsto\:Coordinates \: of \: P = \bigg(\dfrac{12}{3} ,\dfrac{6}{3} \bigg)$

$\bf :\longmapsto\:Coordinates \: of \: P = (4,2)$

Now,

Case :- 2 When Q divides AB in the ratio 2 : 1

So, using Section Formula

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

Here,

$\rm :\longmapsto\:x_1 = 3$

$\rm :\longmapsto\:y_1 = - \: 1$

$\rm :\longmapsto\:x_2= 6 \:$

$\rm :\longmapsto\:y_2= 8 \:$

$\rm :\longmapsto\:m = 2$

$\rm :\longmapsto\:n = 1$

So, on substituting the values,

$\rm :\longmapsto\:Coordinates \: of \: Q = \bigg(\dfrac{2 \times 6 + 1 \times 3}{1 + 2} ,\dfrac{2 \times 8 + 1 \times ( - 1)}{2 + 1}\bigg)$

$\rm :\longmapsto\:Coordinates \: of \: Q = \bigg(\dfrac{12+ 3}{3} ,\dfrac{16 - 1}{3} \bigg)$

$\rm :\longmapsto\:Coordinates \: of \: P = \bigg(\dfrac{15}{3} ,\dfrac{15}{3} \bigg)$

$\bf :\longmapsto\:Coordinates \: of \: Q = (5,5)$

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{Coordinates \: of \: P = (4,2)} \\ \\ &\sf{Coordinates \: of \: Q = (5,5)} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

1. Distance formula is used to find the distance between two given Points.

${\underline{\boxed{\rm{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

2. Mid Point formula is used to find the Mid points on any line.

$\boxed{ \rm{ \: (x,y) = \bigg(\dfrac{x_1 + x_2}{2} ,\dfrac{y_1 + y_2}{2} \bigg)}}$