Question

IN WHAT RATIO DOES THE POINT (-4,6) DIVIDE THE LINE SEGMENT AB JOINING THE POINTS A(-6,10) AND B(3,-8) ?
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Answer 1

Given :-

Point (-4,6)

A (-6,10)

B(3,-8)

To Find :-

Ratio

Solution :-

Using section formula

$\sf (x,y) =\bigg(\dfrac{mx_2+nx_1}{m+n},\dfrac{mx_1+nx_2}{m+n}\bigg)$

$\sf (-4,6)=\bigg(\dfrac{m(3)+n(-6)}{m+n},\dfrac{m(-8)+n(10)}{m+n}\bigg)$

Let

$\sf -4=\dfrac{m(3)+n(-6)}{m+n}(..i)$

$\sf -4(m)+-4(n)=3m+(-6n)$

$\sf -4m-4n=3m-6n$

$\sf -4m-3m=-6n+4n$

$\sf -7m=-2n$

$\sf 7m=2n$

$\sf\dfrac{7}{2}=\dfrac{m}{n}$

$\sf 7:2=m:n$

Answer 2

Answer:

$\bold{m_{1}:m_{2}=2:7\;is\;the\;required\;ratio.}$

Step-by-step explanation:

$\bold{Let\;(-4,6)\;divide\;AB\;internally\;in\;the\;ratio\;m_1:m_2.}$

$\bold{Using\;the\;section\;formuala\;we\;get:}$

$\bold{(-4,6)={\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_1+m_2} ;\dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_1+m_2} } }$

$\bold{-4,6=\dfrac{m_{1}(3)+m_{2}(-6)}{m_{1}+m_{2}};\dfrac{m_{1}(-8)+m_{2}(10)}{m_{1}+m_{2}} }$

$\bold{-4,6=\dfrac{3m_{1}-6m_{2}}{m_{1}+m_{2}};\dfrac{-8m_{1}+10m_{2}}{m_{1}+m_{2}} }$

$\bold{On\;comparing\;L.H.S\;to\;R.H.S\;we\;get:}$

$\bold{-4=\dfrac{3m_{1}-6m_{2}}{m_{1}+m_{2}} }$

$\bold{Cross\;multiplying\;we\;get:}$

$\bold{-4m_{1}-4m_{2}=3m_{1}-6m_{2}}$

$\bold{-4m_{1}-3m_{1}=-6m_{2}+4m_{2}}$

$\bold{-7m_{1}=-2m_{2}}$

$\therefore\bold{m_{1}:m_{2}=2:7}$