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### WB JEE 2008

Subjective

If 2a $$-$$ 5b $$-$$ 3c = 0, show that the straight line ax + by + c = 0 always passes through a fixed point. Find the equation of one straight line passing through this fixed point and parallel to the line 9x + 12y = 20.

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## Explanation

We have, $$2a - 5b - 3c = 0$$.

or, $$c = {{2a - 5b} \over 3}$$ ..... (i)

Hence, equation of the straight line can be written as

$$ax + by + {{2a - 5b} \over 3} = 0$$

or, $$3ax + 3by + 2a - 5b = 0$$

or, $$a(3x + 2) + b(3y - 5) = 0$$.

This will be true for all values of a and b, only if (3x + 2) and (3y $$-$$ 5) are simultaneously zero which means, for x = $$-$$2/3 and y = 5/3, the equation is always satisfied for any values of a and b.

$$\therefore$$ The straight line passes through the point ($$-$$2/3, 5/3)

Now, slope of the given line 9x + 12y = 20 is $$-$$3/4

Any straight line parallel to this line will also have slope of $$-$$3/4

$$\therefore$$ Equation of the line parallel to 9x + 12y = 20 and passing through ($$-$$2/3, 5/3) is

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$ or, $$- 4y + {{20} \over 3} = 3x + 2$$

or, $$3x + 4y - {{14} \over 3} = 0$$ or, $$9x + 12y - 14 = 0$$.

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