The line parallel to x-axis and passing through the point of intersection of lines ax+2by+…

Question

The line parallel to $x$-axis and passing through the point of intersection of lines $ax+2by+3b=0$ and $bx-2ay-3a=0$ where $(a,b)
e(0,0)$ is:

Accepted Answer

The line passing through the intersection of lines

$ax + 2by = 3b = 0$

and $bx - 2ay - 3a = 0$ is

$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$

As this line is parallel to $x$-axis.

$$\therefore$$ $$a + b\lambda  = 0 \Rightarrow \lambda  =  - a/b$$

$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$

$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$

$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$

$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) =  - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$

$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$

So it is $3/2$ units below $x$-axis.

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