Question

find the coordinates of the points of trisection of the line segment joining the points A(-5,6) and B(4-3)​

Answer 1

Answer:

$\bigstar{\bold{Coordinates\:of\:P=(-2,3)}}$

$\bigstar{\bold{Coordinates\:of\:Q=(1,0)}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• The point A (-5, 6)
• The point B (4,-3)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Coordinates of the point of trisection

$\Large{\underline{\underline{\bf{Solution:}}}}$

➣ Let the points P and Q trisect the line.

➣ Since P and Q trisects the line, P would divide line segment AB in the ratio 1 : 2

➣ Hence we can find the coordinates of P

➣ By distance formula,

    $\sf{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$

    where m₁ = 1, m₂ = 2, x₁ = -5, x₂ = 4, y₁ = 6, y₂ = -3

➣ Substituting the datas,

    $\sf{(x,y)=\bigg(\dfrac{4-10}{1+2} ,\dfrac{-3+12}{1+2}\bigg) }$

    $\sf{(x,y)=\bigg(\dfrac{-6}{3} ,\dfrac{9}{3}\bigg) }$

    $\sf{(x,y)=(-2,3)}$

➣ Hence the coordinates of P are (-2 , 3)

    $\boxed{\bold{Coordinates\:of\:P=(-2,3)}}$

➣ Now we have to find the coordinates of Q

➣ Q divides line segment AB in the ratio 2 : 1

➣ By distance formula,

    $\sf{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$

    where m₁ = 2, m₂ = 1, x₁ = -5, x₂ = 4, y₁ = 6, y₂ = -3

➣ Substitute the datas,

    $\sf{(x,y)=\bigg(\dfrac{8-5}{3},\dfrac{-6+6}{3} \bigg)}$

    $\sf{(x,y)=\bigg(\dfrac{3}{3},\dfrac{0}{3} \bigg)}$

     $\sf{(x,y)=(1,0)}$

➣ Hence the coordinates of Q is (1,0)

    $\boxed{\bold{Coordinates\:of\:Q=(1,0)}}$

       $\setlength{\unitlength}{1 cm}\begin{picture}(6,6)\put(3,2){\vector(1,0){3}}\put(3,2){\vector(-1,0){3}}\put(2,1.96){\circle*{0.15}}\put(2,1.5){\large\sf P}\put(4,1.96){\circle*{0.15}}\put(4,1.5){\large \sf Q}\put(0,1.5){\large \sf A(-5,6)}\put(5.8,1.5){\large \sf B(4,-3)}\end{picture}$

   

$\Large{\underline{\underline{\bf{Notes:}}}}$

➣ The distance formula is given by,

    $\sf{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$