Some more examples of pair of straight lines questions for IITJEE math

Problem 1:

If one of the straight lines given by the equation $ax^{2}+2hxy+by^{2}=0$ coincides with one of those given by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$, and the other lines represented by them be perpendicular, prove that $\frac{ha^{'}b^{'}}{b^{'}-a^{'}}=\frac{h^{'}ab}{b-a}=\frac{1}{2}\sqrt{aa^{'}bb^{'}}$

Solution 1:

Let the lines represented by $ax^{2}+2hxy+by^{2}=0$ be $y=m_{1}x$ and $y=m_{2}x$, so that

$m_{1}+m_{2}=\frac{2h}{b}$ and $m_{1}m_{2}=\frac{a}{b}$…call this equation (A)

The lines represented by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$ are $y=m_{1}x$ and $y=-\frac{1}{m_{2}}x_{1}$ so that

$m_{1}-\frac{1}{m_{2}}=-\frac{2h^{'}}{b^{'}}$ and $m_{1}(-\frac{1}{m_{2}})=\frac{a^{'}}{b^{'}}$…call this equation (B)

From (A) and (B), we get $m_{1}m_{2}(-\frac{m_{1}}{m_{2}})=\frac{aa^{'}}{bb^{'}}$

which in turn $\Longrightarrow m_{1}^{2}=\frac{aa^{'}}{bb^{'}}$,

which in turn $\Longrightarrow m_{1}=\sqrt{(-\frac{aa^{'}}{bb^{'}})}$

and from (A), again, we get $\sqrt{-\frac{aa^{'}}{bb^{'}}}-\sqrt{-\frac{aa^{'}}{ba^{'}}}=-\frac{2h}{b}$

and this in turn $\Longrightarrow \sqrt{(-aa^{'}bb^{'})}(\frac{1}{bb^{'}}-\frac{1}{a^{'}b})=-\frac{2h}{b}$, that is, it $\Longrightarrow \frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{ha^{'}b^{'}}{b^{'}-a^{'}}$.

Similarly, from (B), we get $\frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{h^{'}ab}{b-a}$. Hence, the required result follows.

Problem 2:

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents two straight lines, prove that the equation of the third pair of straight lines passing through the points where these meet the axes $xy=0$ is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$.

Solution 2:

Equation of the lines passing through the intersection of  the given lines and axes is

$ax^{2}+2hxy+by^{2}+2gx+2fy+c+2xy \lambda =0$, or

$ax^{2}+2(h+\lambda)xy+by^{2}+2gx+2fy+c=0$

Since it represents a pair of straight lines

$abc+2fg(h+\lambda)-af^{2}-bg^{2}-c(h+\lambda)^{2}=0$,

$\Longrightarrow abc+2fgh-af^{2}-bg^{2}-ch^{2}+2fg\lambda -2ch\lambda -c\lambda^{2}=0$

$\Longrightarrow \lambda = \frac{2(fg-ch)}{c}$ (since $abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$).

Hence, the required equation of the lines is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$

Problem 3:

Show that the equation $\sqrt{3}x^{2}-(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}-\sqrt{3}y^{2}=0$ represents three straight lines through the origin such that one of them bisects the angle between the other two. Also find the equation of  the lines perpendicular to the given lines through the origin.

Solution 3:

The given equation can be written as $\sqrt{3}(x^{3}-y^{3})-(4+\sqrt{3})xy(x-y)=0$, or

$(x-y)(\sqrt{3}(x^{2}+xy+y^{2}))-(4+\sqrt{3})xy=0$, or

$(x-y)(\sqrt{3}x^{2}-4xy+\sqrt{3}y^{2})=0$, or

$(x-y)(\sqrt{3}x-y)(x-\sqrt{3}y)=0$,or

$x-y=0$, $\sqrt{3}x-y=0$, and $x-\sqrt{3}y=0$

which gives three straight lines passing through the origin with slopes 45 degrees, 60 degrees, and 30 degrees respectively showing that $x-y=0$ bisects the angles between the other two.

Now, equations of the lines through the origin perpendicular to these lines are

$x+y=0$, $x+\sqrt{3}y=0$, $\sqrt{3}x+y=0$ so their joint equation is given by

$(x+y)(x+\sqrt{3}y)(\sqrt{3}x+y)=0$

or, $\sqrt{3}x^{3}+(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}+\sqrt{3}y^{3}=0$

That’s all, folks !

Nalin Pithwa.

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