Question

find the points of trisection of the line se
gment formed by (-3,-5),(-6,-8)​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

Find the points of trisection of line se gment formed by ( -3 , -5) and ( -6 , -8)

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{P\:=\:\left(-4,-6\right)}}\boxed{\sf{Q\:=\:\left(-5,-7\right)}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

m : n = 1 : 2

m = 1

n = 2

(-3, -5) , (-6, -8)

 x₁   y₁     x₂   y₂

$\large\boxed{\sf{P\:=\:\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}}$

$\sf{P\:=\:\left(\dfrac{1(-6)+2(-3)}{1+2},\dfrac{1(-8)+2(-5)}{1+2}\right)}$

$\sf{P\:=\:\left(\dfrac{-6+(-6)}{3},\dfrac{-8+(-10)}{3}\right)}$

$\sf{P\:=\:\left(\dfrac{-6-6}{3},\dfrac{-8-10}{3}\right)}$

$\sf{P\:=\:\left(\dfrac{-12}{3},\dfrac{-18}{3}\right)}$

$\boxed{\sf{P\:=\:\left(-4,-6\right)}}$

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m : n = 2 : 1

m = 2

n = 1

(-3, -5) , (-6, -8)

 x₁   y₁     x₂   y₂

$\large\boxed{\sf{Q\:=\:\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}}$

$\sf{Q\:=\:\left(\dfrac{2(-6)+1(-3)}{2+1}, \dfrac{(-8)+1(-5)}{2+1}\right)}$

$\sf{Q\:=\:\left(\dfrac{-12 -3}{3}, \dfrac{-16 - 5}{3}\right)}$

$\sf{Q\:=\:\left(\dfrac{-15}{3}, \dfrac{-21}{3}\right)}$

$\boxed{\sf{Q\:=\:\left(-5,-7\right)}}$