Chapter 6:Coordinate Geometry Given that A(6,8),B(p,-3) and C(-4,-6) are three points lying on a straight line such that AB:BC=m:n.Find a)the ratio m:n b)the value of p
@mathmath333
\(\large \color{black}{\begin{align}& \normalsize \text{For finding 'p' u need to equate the slope .}\hspace{.33em}\\~\\ &\normalsize \text{as the three points are collinear AB and BC should have same slope.}\hspace{.33em}\\~\\ &\dfrac{8-(-3)}{6-p}=\dfrac{-3-(-6)}{p-(-4)} \end{align}}\)
okay
\(\large \color{black}{\begin{align} \normalsize \text{u can find m:n only after finding 'p'}\hspace{.33em}\\~\\ \end{align}}\)
\[p=-\frac{ 13 }{ 7 }\]
correct
For finding \(m:n\) u need the section formula
|dw:1431529356233:dw|
okay
\(\large \color{black}{\begin{align} x=\dfrac{nx_1+mx_2}{m+n}\hspace{.33em}\\~\\ \end{align}}\)
put the values and get the answer
\[-\frac{ 13 }{ 7 }=\frac{ 6n-4m }{ m+n }\]
\(A(6,8),B(p,-3) ~and~ C(-4,-6)\\ A(x_1,y_1),B(x,y)~and~C(x_2,y_2)\)
\[-\frac{ 13 }{ 7 }=\frac{ 6n-4m }{ m+n }\]
yea simplify it
\[-13m-13n=42n-28m\]
\[15m-55n=0\]@mathmath333
@mathmath333
Am i correct? @mathmath333
yes
what should we do next? @mathmath333
find \(\dfrac{m}{n}\)
i'm not sure how to do,can u show me pls?
\(\large \color{black}{\begin{align} 15m-55n=0\hspace{.33em}\\~\\ 3m-11n=0\hspace{.33em}\\~\\ 3m=11n\hspace{.33em}\\~\\ \dfrac{m}{n}=\dfrac{11}{3} \end{align}}\)
Thank you @mathmath333 :)
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