Question

Find the coordinates of the points of trisection of the line segment joining the points (1, -2) and (-3, 4).​

Answer 1

Step-by-step explanation:

Concept used:-

The coordinates of the point P( x , y) divides the line segment joining the points A( x₁ , y₁) and B(x₂ , y₂) internally in the ratio  m₁ : m₂ are

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\boxed{ \sf \: \bigg(\frac{ m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \: , \: \frac{ m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}\bigg)}$

Solution:- ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀

Let P and Q be the points of trisection of the line joining the points (1, -2) and (-3, 4).

Then, AP = PQ = QB

As, P and Q are the points of trisection.

P divides AB internally in the ratio 1 : 2 and

Here, m₁ = 1 and m₂ = 2

• x₁ = 1 , x₂ = -3 , y₁ = -2 and y₂ = 4

By applying section formula

$\sf The \: coordinates \: of \: P = \: \bigg(\dfrac{1( - 3) + 2(1)}{1 + 2} \: , \: \dfrac{ 1(4) + 2( - 2)}{1 + 2}\bigg)$

$\sf = \: \bigg( \dfrac{ - 3 + 2}{3} \: , \: \dfrac{4 - 4}{3}\bigg)$

$\sf = \: \dfrac{ - 1}{3} \: , \: 0$

$\sf \therefore \underline{\:\:{\underline {\: The \: coorinates \: of \: P\: are \: \bigg(\dfrac{ - 1}{3} \: , \: 0 \bigg)}\:}\:\:}$

Now, Q divides AB internally in the ratio 2 : 1.

Here, m₁ = 2 and m₂ = 1

• x₁ = 1 , x₂ = -3 , y₁ = -2 and y₂ = 4

By applying section formula

$\sf The \: coordinates \: of \: Q = \: \bigg(\dfrac{2( - 3) + 1(1)}{1 + 2} \: , \: \dfrac{ 2(4) + 1( - 2)}{1 + 2}\bigg)$

$\sf = \: \bigg( \dfrac{ - 6 + 1}{3} \: , \: \dfrac{8- 2}{3}\bigg)$

$\sf = \: \dfrac{ - 5}{3} \: , \: 2$

$\sf \therefore \underline{\:\:{\underline {\: The \: coorinates \: of \: Q\: are \: \bigg(\dfrac{ - 5}{3} \: , \: 2 \bigg)}\:}\:\:}$

Therefore:-

The coorinates of the points of trisection of the line segment joining the points (1, -2) and (-3 , 4) are are (-1/3 , 0) and (-5/3 , 2)

Answer 2

What is trisection?

Refer the attachment-In attachment you will see that AC=CD=DC

This means

AC:CD=1:1

CD:DB=1:1

AC:CB=1:2

AD:DB=2:1

What we have to find?

We have to find the coordinates of points C and D

First we find the coordinates of C

Using section formula

$\rm(x,y)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}) \\ \implies \rm C(x,y) = ( \frac{1 \times - 3 + 2 \times 1}{3} \: , \frac{1 \times 4 + 2 \times - 2 }{3} \\ \implies \boxed{ \rm C(x,y) = (\frac{ - 1}{3} 0) }$

Now we find the coordinates of D

Using mid point formula

$\rm \: (x,y) = ( \frac{x1 + x2}{2}, \frac{y1 + y2}{2}) \\ \rm \implies D(x,y) = ( \frac{ - \frac{1}{3} - 3}{2}, \frac{ 0+ 4}{2}) \\ \rm \boxed {D(x,y) = ( \frac{ - 5}{3} ,{2})}$