Question

Find the ratio in which the point P(3/4,5/12) divides the line segment joining the points A(-1/2,3/2) and B(2,-5).
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Answer 1

Question:

Find the ratio in which the point P (3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B (2,-5)

Answer:

$\boxed{\bold{Ratio=1:5}}$

Step-by-step explanation:

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

• Point A = (1/2, 3/2)
• Point B = (2,-5)
• Point P = (3/4,5/12)

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

• The ratio in which point P divides the line segment AB

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

→ Let us assume point P divides the line segment in the ratio k : 1

→ We know by section 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 x = 3/4, y = 5/12 , x₁ = 1/2, x₂ = 2, y₁ = 3/2, y₂ = -5, m₁ = k, m₂ = 1

→ Substituting the values we get,

   $\sf{\bigg(\dfrac{3}{4},\dfrac{5}{12} \bigg)=\bigg(\dfrac{2k+\frac{1}{2} }{k+1},\dfrac{-5k+\frac{3}{2} }{k+1}\bigg)}$

→ Simplifying we get,

  $\sf{\bigg(\dfrac{3}{4},\dfrac{5}{12} \bigg)=\bigg(\dfrac{\frac{4k+1}{2} }{k+1},\dfrac{\frac{-10k+3}{2} }{k+1}\bigg)}$

  $\sf{\bigg(\dfrac{3}{4},\dfrac{5}{12} \bigg)=\bigg(\dfrac{4k+1 }{2k+2},\dfrac{-10k+3 }{2k+2}\bigg)}$

→ Equating it we get,

  $\sf{\dfrac{3}{4}=\dfrac{4k+1}{2k+2}}$

→ Cross multiplying,

  3 ( 2k + 2) = 4 (4k + 1)

  6k + 6 = 16k + 4

  6k - 16k = 4 - 6

  -10k = -2

     k = -2/-10

     k = 1/5

→ Equating the y coordinate,

  $\sf{\dfrac{5}{12}=\dfrac{-10k+3}{2k+2}}$

→ Cross multiplying,

  5 (2k + 2) = 12 (-10k + 3)

  10k + 10 = -120k + 36

  10k + 120k = 36 - 10

  130k = 26

        k = 26/130

        k = 1/5

→ Therefore the line is divided in the ratio 1 : 5

   $\boxed{\bold{Ratio=1:5}}$

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

→ The section formula is given by

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