Question

if the points (a,0), (0,b), (3,2) are collinear. prove that 2/b + 3/a = 1

Answer 1

Answer:

Step-by-step explanation:

Please refer the attachment below

Thank you....

Answer 2

Answer:

Given (a,0), (0,b) and (3,2) are collinear.

$Slope \: of \: the\:line \\containing \:the \:points \\(x_{1},y_{1}),\:and\:(x_{2},y_{2}),\:is \:m =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$Slope \:of \:the \:line\: segment\:\\joining \:points (a,0)\:and\:(0,b)\\=\frac{b-0}{0-a}\\=\frac{-b}{a}\:--(1)$

$Slope \:of \:the \:line\: segment\:\\joining \:points (0,b)\:and\:(3,2)\\=\frac{2-b}{3-0}\\=\frac{2-b}{3}\:--(2)$

/* From (1) and (2),

$\frac{-b}{a}=\frac{2-b}{3}\:(points \:are \: collinear)$

$\implies -3b = a(2-b)$

$\implies -3b = 2a-ab$

$\implies -2a-3b = -ab$

/* Divide each term by -ab ,we get

$\implies \frac{-2a}{-ab}-\frac{3b}{-ab}=\frac{-ab}{-ab}$

$\implies \frac{2}{b}+\frac{3}{a}=\frac{1}{1}$

$\implies \frac{2}{b}+\frac{3}{a}=1$

Therefore,

$\implies \frac{2}{b}+\frac{3}{a}=1$

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