Question

a line segment joining two points (-3,-4) and (1,-2) is devide by y-axis in what ratio??



plz guys if u know correct answer than answer it pls....​

Answer 1

Step-by-step explanation:

Let m : n be the required ratio. The point that divides the line segment joining the points (-3, -4) and (1, -2) is written as (0, t). Thus, the required ratio is 3 : 1

Answer 2

Answer:

$\bigstar{\bold{Ratio=3:1}}$

Step-by-step explanation:

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

• Point A = (-3, - 4)
• Point B = (1, - 2)
• The line segment is divided by the y axis

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

• The ratio in which the line segment is divided

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

→ Let the line be divided in the ratio k : 1

→ Here it is given that the line segment is divided by the y axis.

→ Hence the coordinates of division = (0,y)

→ By section formula we know that,

  $\sf{(0,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2} ,\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \bigg)}$

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

→ Substituting the datas we get,

   $\sf{(0,y)=\bigg(\dfrac{k-3}{k+1} ,\dfrac{-2k-4}{k+1} \bigg)}$

→ Equating it,

  $\sf{\dfrac{k-3}{k+1}=0}$

→ Cross multiplying it,

   k - 3 = 0

   k = 3

→ But we know that line segment is divided in the ratio k : 1

→ Hence the line segment is divided in the ratio 3 : 1

$\boxed{\bold{Ratio=3:1}}$

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

→ The section formula is given by,

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