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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