Question

Find the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).​

Answer 1

Answer:

m₁ : m₂ = 2 : 1

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Step-by-step explanation:

Let us name the points (15, 5) and (9, 20), A and B respectively. The point (11, 15), let us name it C, intersects the line AB, and we're asked to find in which ratio the point C divides the line AB.

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For any two points A(x₁, y₁), B(x₂, y₂) being divided by a point C(x, y), the coordinates of C(x₃, y₃) are given by the Section formula.

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

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Where m1 : m2 refers to the ratio in which the point divides the line AB and x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂) .

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We'll be using this formula to find out the ratio in which AB is being divided by the point C(11, 15).

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Here;

• x₁ = 15

• x₂ = 9

• y₁ = 5

• y₂ = 20

• x = 11

• y = 15

• m₁ = ?

• m₂ = ?

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$\dashrightarrow \ \sf C(11, 15) = \Bigg\{\dfrac{m_1(9) + m_2(15)}{m_1 + m_2} , \dfrac{m_1(20) + m_2(5)}{m_1 + m_2}\Bigg\}$

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$\dashrightarrow \ \sf C(11, 15) = \Bigg\{\dfrac{9m_1 + 15m_2}{m_1 + m_2} , \dfrac{20m_1 + 5m_2}{m_1 + m_2}\Bigg\}$

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On comparing x with (m₁x₂ + m₂x₁)/(m₁ + m₂) we get;

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$\dashrightarrow \ \sf 11 = \dfrac{9m_1 + 15m_2}{m_1 + m_2}$

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$\dashrightarrow \ \sf 9m_1 + 15m_2 = 11(m_1 + m_2)$

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$\dashrightarrow \ \sf 9m_1 + 15m_2 = 11m_1 + 11m_2$

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$\dashrightarrow \ \sf 15m_2 - 11m_2 = 11m_1 - 9m_1$

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$\dashrightarrow \ \sf 4m_2 = 2m_1$

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$\sf \dashrightarrow \ \dfrac{4}{2} = \dfrac{m_1}{m_2}$

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$\dashrightarrow \ \boxed{\sf \dfrac{2}{1} = \dfrac{m_1}{m_2}}$

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On comparing y with (m₁y₂ + m₂y₁)/(m₁ + m₂) we get;

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$\sf \dashrightarrow 15 = \dfrac{20m_1 + 5m_2}{m_1 + m_2}$

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$\sf \dashrightarrow 15(m_1 + m_2) = 20m_1 + 5m_2$

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$\sf \dashrightarrow 15m_1 + 15m_2 = 20m_1 + 5m_2$

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$\sf \dashrightarrow 15m_1 - 20m_1 = 5m_2 - 15m_2$

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$\sf \dashrightarrow -5m_1 = -10m_2$

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$\sf \dashrightarrow \dfrac{m_1}{m_2} = \dfrac{-10}{-5}$

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$\sf \dashrightarrow \boxed{\sf \dfrac{m_1}{m_2} = \dfrac{2}{1}}$

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∴ The line joining the points (15, 5) and (9, 20) is divided in the ratio 2 : 1 by the point (11, 15).

Answer 2

Step-by-step explanation:

Let mark the points as P,Q,R

Given:-

• P(15,5)
• Q(9,20)
• R(11,15)

To find:-

The adjoining ratio or m:n=?

Solution:-

$\sf we\:have \begin{cases}\sf x_1=15 \\ \sf x_2=9 \\ \sf y_1=5 \\ \sf y_2=20 \\ \sf x=11 \\ \sf y=15\end{cases}$

We know that

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

$\\ \sf{:}\rightarrowtail (11,15)=\left(\dfrac{m(9)+n(15)}{m+n},\dfrac{m(20)+n(5)}{m+n}\right)$

$\\ \sf{:}\rightarrowtail (11,15)=\left(\dfrac{9m+15n}{m+n},\dfrac{20m+5n}{m+n}\right)$

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$\\ \sf{:}\rightarrowtail \dfrac{9m+15n}{m+n}=11$

$\\ \sf{:}\rightarrowtail 9m+15n=11(m+n)$

$\\ \sf{:}\rightarrowtail 9m+15n=11m+11n$

$\\ \sf{:}\rightarrowtail 15n-11n=11m-9n$

$\\ \sf{:}\rightarrowtail 4n=2m$

$\\ \sf{:}\rightarrowtail \dfrac{m}{n}=\dfrac{4}{2}$

$\\ \bf{:}\rightarrowtail \dfrac{m}{n}=\dfrac{2}{1}$

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$\\ \sf{:}\rightarrowtail \dfrac{20m+5n}{m+n}=15$

$\\ \sf{:}\rightarrowtail 20m+5n=15(m+n)$

$\\ \sf{:}\rightarrowtail 20m+5n=15m+15n$

$\\ \sf{:}\rightarrowtail 20m-15m=15n-5n$

$\\ \sf{:}\rightarrowtail 5m=10n$

$\\ \sf{:}\rightarrowtail \dfrac{m}{n}=\dfrac{10}{2}$

$\\ \bf{:}\rightarrowtail \dfrac{m}{n}=\dfrac{2}{1}$

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• We got

$\\ \sf{:}\rightarrowtail \dfrac{m}{n}=\dfrac{2}{1}$

$\\ {:}\rightarrowtail \boxed{\mathfrak{m:n=2:1}}$