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2015-02-09T19:24:02+05:30

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If a point (x,y) lies on a line joining the points A(x₁,y₁) and B(x₂,y₂), the equation of the line is given by

 \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}

Point P(x,y) lies on the line joining the points A(a,0) and B(0,b). So

 \frac{y-0}{x-a} = \frac{b-0}{0-a} \\ \\ \Rightarrow \frac{y}{x-a}=- \frac{b}{a} \\ \\ \Rightarrow ay=-b(x-a)\\ \\ \Rightarrow ay=-bx+ab\\ \\ \Rightarrow ay+bx=ab\\ \\divide\ both\ sides\ by\ ab\\ \\ \Rightarrow  \frac{ay}{ab}+ \frac{bx}{ab} = \frac{ab}{ab}\\ \\ \Rightarrow  \frac{y}{b}+ \frac{x}{a}  =1\ (proved)



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The Brainliest Answer!
2015-02-09T19:27:21+05:30
Let the ratio in which P(x,y) divides AB be k:1

x=a+0/k+1
y=0+bk/k+1
 
LHS
 \frac{x}{a} + \frac{y}{b} =  \frac{\frac{a}{k+1}}{a}  + \frac{ \frac{bk}{k+1} }{b} = \frac{a}{a(k+1)} + \frac{kb}{b(k+1)} = \frac{1}{k+1}+ \frac{k}{k+1} = \frac{1+k}{k+1}  =1 RHS
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