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Question

Find the ratio in which the line segment joining the points $$(-3,10)$$ and $$(6-8)$$ is divided by $$(-1,6)$$


Solution

Consider the points be $$A\left( { - 3,10} \right),B\left( {6, - 8} \right),C\left( { - 1,6} \right)$$

And we need to find the ratio between $$AC$$ and $$CB$$

So, let the ratio be $$k:1$$

Therefore, 

$${m_1} = k,{m_2} = 1$$

$$ { x_{ 1 } }=-3,{ y_{ 1 } }=10 $$

$$ { x_{ 2 } }=6,{ y_{ 2 } }=-8$$

And $$x =  - 1,y = 6$$

Now, using section formula 

$$ x=\dfrac { { { m_{ 1 } }{ x_{ 2 } }+{ m_{ 2 } }{ x_{ 1 } } } }{ { { m_{ 1 } }+{ m_{ 2 } } } }  $$

$$ \Rightarrow -1=\dfrac { { k\times 6+1\times -3 } }{ { k+1 } }  $$

$$ \Rightarrow -1=\dfrac { { 6k-1 } }{ { k+1 } }  $$

$$ \Rightarrow -1\left( { k+1 } \right) =6k-3 $$

$$ \Rightarrow -k-1=6k-3 $$

$$ \Rightarrow -k-6k=-3+1 $$

$$ \Rightarrow -7k=-2 $$

$$ \Rightarrow k=\dfrac { { -2 } }{ 7 }  $$

$$ \Rightarrow k=\dfrac { 2 }{ 7 }  $$

Therefore, the ratio is $$k:1$$

$$ = \dfrac{2}{7}:1$$ 

Now multiply $$7$$ both sides

$$ =7\times \dfrac { 2 }{ 7 } :7\times 1 $$

$$ =2:7 $$

Hence, The Ratio is $$2:7$$

Mathematics

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