Show that the points (a, 0), (0, b) and (3a, -2b) are collinear. Also find the equation of the line containing them.

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Solution

Let A(a, 0), B(0, b) and C(3a, -2b) and C(3a, -2b) be the given points.

Then, the equation of line AB is given by $\frac{y - 0}{x - a} = \frac{b - 0}{0 - a} \Rightarrow - a y = b x - a b$

$\Rightarrow b x + a y - a b = 0$

Thus, the equation of line AB is $b x + a y - a b = 0$ .............(i)

Putting x = 3a and $y = - 2 b$ in (i), we get LHS =

$b . 3 a + a (- 2 b) - a b = 3 a b - 2 a b - a b = 0 = RHS$

Thus, the point $C (3 a, - 2 b)$ also lies on AB.

Hence, the given points are collinear and the equation of the line containing them is $b x + a y - a b = 0$

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