Question

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

A
1:7
B
2:7
C
2:5
D
3:7

Solution

## The correct option is B $$2:7$$Using the section formula, if a point $$(x,y)$$ divides the line joining the points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$\displaystyle ({ x }_{ 2 },{ y }_{ 2 })$$ in the ratio $$m:n$$, then $$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$$Let the ratio be $$k : 1$$Substituting $$({ x }_{ 1 },{ y }_{1 }) = (-3,10)$$ and $$({x }_{ 2 },{ y }_{ 2 }) = (6,-8)$$  in the sectionformula, we get  $$\displaystyle \left( \frac { k(6) + 1(-3) }{ k + 1 },\frac { k(-8) + 1(10) }{ k + 1 } \right) = ( -1, 6)$$ $$\displaystyle \left( \frac { 6k - 3 }{ k + 1 } ,\frac { -8k + 10 }{ k + 1} \right) = ( -1,6)$$Comparing the x - coordinate,$$\displaystyle \frac { 6k-3 }{ k+1 } =-1$$ $$\implies 6k - 3 = -k - 1$$$$7k = 2$$$$\displaystyle k = \frac {2}{7}$$Hence, the ratio is $$2:7$$Mathematics

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