Question

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

Answer 1

Given :-

Two points A (3,-2) and B (-3,-4)

To find :-

Point of trisection of the line segment joining points (3,-2) and (-3,-4)

Knowledge required :-

• section formula

$\green{\bigstar}\boxed{\sf{(x,y)=\left( \dfrac{mx_2+nx_1}{m+n}\right) , \left( \dfrac{my_2+ny_1}{m+n}\right)}}$

where (x,y) tells us the coordinates of point which divide the line segment joining points (x₁ , y₁) and (x₂ , y₂) in the ratio m : n .

Figure :-

$\setlength{\unitlength}{1cm}\begin{picture}(7,7)\linethickness{0.3mm}\put(0,0){\line(1,0){6}}\put(2,0){\circle*{0.1}}\put(4,0){\circle*{0.1}}\put(-0.5,-0.5){\bf{A\;(3,-2)}}\put(1.7,-0.5){\bf{C\;(p,q)}}\put(3.7,-0.5){\bf{D\;(r,s)}}\put(6,-0.5){\bf{B\;(-3,-4)}}\put(1,0.5){\bf{1\;\;\;\;\;\;: \;\;\;\;\;1 \;\;\;\;\;\;\;\;:\;\;\;\;\;\;1 }}\end{picture}$

• A (3,-2) and B(-3,-4) are coordinates of points making line segment.
• C (p,q) and D (r,s) are coordinates of point of trisection of line segment AB.

Solution :-

• coordinates of point A are (3,-2)
• coordinated of point B are (-3,-4)
• coordinates of point C are (p,q)
• AC : CB = 1 : 2

Using section formula

$\longmapsto\sf{(p,q)=\left(\dfrac{(1)(-3)+(2)(3)}{1+2},\dfrac{(1)(-4)+(2)(-2)}{1+2}\right)}\\ \\ \\ \longmapsto\sf{(p,q)=\left( \dfrac{-3+6}{3},\dfrac{-4-4}{3}\right)}\\\\\\\longmapsto\boxed{\red{\sf{(p,q)=\left(1,\dfrac{-8}{3}\right)}}}$

Now,

• coordinates of point C are (1,-8/3)
• coordinates of point B are (-3,-4)
• coordinates of point D are (r,s)
• CD : DB = 1 : 1

Using section formula

$\longmapsto\sf{(r,s)=\left(\dfrac{(1)(-3)+(1)(1)}{1+1},\dfrac{(1)(-4)+(1)(\dfrac{-8}{3})}{1+1}\right)}\\\\\\\longmapsto\sf{(r,s)=\left(\dfrac{-3+1}{2},\dfrac{-4-\dfrac{8}{3}}{2}\right)}\\\\\\\longmapsto\boxed{\red{\sf{(r,s)=\left(-1,\dfrac{-20}{6}\right)}}}$

So,

Points of trisection of line segment joining points (3,-2) and (-3,-4) are (1,-8/3) and (-1,-20/6) .