Any two lines through the Origin may be written as y = mx and y = tx where m and t are their gradients(slopes). So (y - mx)(y - tx) = 0 giving y - mx or y - tx = 0 must represent the pair.
The general form of this equation is given by:
$$ax^2+2hxy+by^2=0$$
which represents the equation of pair of straight lines, simply known as the homogeneous equation of degree two or a second degree homogeneous equation. This equation represents the pair of straight lines passing through an origin. so the general equation of second degree in x and y is of the form,
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$
The homogeneous equation of second degree in x and y is given by,
$$ax^2+2hxy+by^2=0..................(i)$$
Now,
Case I: if b≠0 then,
Dividing equation (i) by bx^{2 } ,
Then , equation (i) becomes,
$$\frac{a}{b}+\frac{2h}{b}(\frac{y}{x})+\frac{y^2}{x^2}=0$$
or,$$(\frac{y}{x})^2+\frac{2h}{b}(\frac{y}{x})+\frac{a}{b}=0............(ii)$$
This equation (ii) is quadratic in y/x. so, it gives two values of y/x,
say that the values are m_{1} and m_{2} then,
$$\frac{y}{x}=m_1$$
$$or,y=m_1x...................(iii)$$
And,
$$\frac{y}{x}=m_2$$
$$or,y=m_2x...................(iv)$$
Hence, both of these equations i.e, equation (iii) and (iv) represents the line through origin.
Case II: if b=0 then,
equation (i) becomes,
$$ax^2+2hxy=0$$
$$or,x(ax+2hy)=0$$
Now, $$Either, x=0$$
$$Or, ax+2hy=0$$
Both of these lines also represents the lines through origin.
Hence, equation (i) always represents the pair of lines through an origin.
Solution,
Let two lines represented by ax^{2}+2hxy+by^{2}=0 are,
$$y=m_1x...................(i)$$
$$y=m_2x...................(ii)$$
Where,
$$m_1+m_2=\frac{-2h}{b}............(iii)$$
$$And,m_1m_2=\frac{a}{b}............(iv)$$
we write equation (iii) and (iv) according to refrence of polynomial equation,
where,$$sum=\frac{-b}{a}$$ and $$product=\frac{c}{a}$$
Now,
LetΘ be the angle between two lines. Then,
$$\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}$$
$$or,\tan\theta=\pm\frac{\sqrt{(m_1+m_2)^2-4m_1m_2}}{1+m_1m_2}$$
$$or,\tan\theta=\pm\frac{\sqrt{\frac{4h^2}{b^2}-4\frac{a}{b}}}{1+\frac{a}{b}}$$
$$or,\tan\theta=\pm\frac{\sqrt{\frac{4h^2-4ab}{b^2}}}{\frac{b+a}{b}}$$
$$\therefore\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
Which is required angle between the line pair represented byax^{2}+2hxy+by^{2}=0.
Case (i): Condition of coincidence i.e,Θ=0°
$$\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$or,\tan0°=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$or,0=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$or,0=\sqrt{h^2-ab}$$
$$or,0=h^2-ab$$
$$\therefore h^2=ab$$
Case (ii): Condition of perpendicular i.e,Θ=90°
$$\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$or,\tan90°=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$or,\frac{1}{0}=\pm\frac{2\sqrt{h^2-ab}}{a+b}$$
$$\therefore a+b=0$$
Note: The homogeneous equation of second degree i.e, ax^{2}+2hxy+by^{2}=0 represents,
The general second degree equation is,
$$ax^2+2hxy+by^2+2gx+2fy+c=0..................(i)$$
The equation (i) can be writen as,
$$ax^2+(2hy+2g)x+(by^2+2fy+c)=0$$
Now, solving for x by quadratic formula, we get,
$$x=\frac{-b\pm\sqrt{h^2-4ac}}{2a}$$
$$or,x=\frac{-(2hy+2g)\pm\sqrt{(2hy+2g)^2-4a(by^2+2fy+c)}}{2a}$$
$$or,x=\frac{-(hy+g)\pm\sqrt{h^2y^2+2ghy+g^2-aby^2-2afy-ca}}{a}$$
Now,
Equation (i) anly represents a pair of straight lines if
h^{2}y^{2}+2ghy+g^{2}-aby^{2}-2afy-ca is a perfect square.
i.e. (h^{2}-ab)y^{2}+(2gh-2af)y+(g^{2}-ca) is a perfect square.
Its discriminant i.e. (B^{2}-4AC) is
$$B^2-4AC=(2gh-2af)^2-4(h^2-ab)(g^2-ca)$$
Now,
Equation (i) represents a pair of straight lines if,
$$(2gh-2af)^2-4(h^2-ab)(g^2-ca)=0$$
solving the above equation, we get,
$$abc+2fgh-af^2-bg^2-ch^2=0$$
which is the required condition for the general equation of second degree to represent a line pair.
The two lines represented by ax^{2}+2hxy+by^{2}=0 are given by,
$$m_1x-y=0.............(i)$$ and,
$$m_2x-y=0...................(ii)$$
Where,$$m_1+m_2=\frac{-2h}{b}$$and $$m_1m_2=\frac{a}{b}$$
The equation of angle bisectors of line (i) and (ii) are given by,
$$\frac{m_1x-y}{\sqrt{1+m_1^2}}=\pm\frac{m_2x-y}{\sqrt{1+m_2^2}}$$
Now,
separating equation of the angle bissectors, we get,
$$\frac{m_1x-y}{\sqrt{1+m_1^2}}-\frac{m_2x-y}{\sqrt{1+m_2^2}}=0$$ and
$$\frac{m_1x-y}{\sqrt{1+m_1^2}}+\frac{m_2x-y}{\sqrt{1+m_2^2}}=0$$
The combined equation of these two angle bisectors is given as,
$$[\frac{m_1x-y}{\sqrt{1+m_1^2}}-\frac{m_2x-y}{\sqrt{1+m_2^2}}][\frac{m_1x-y}{\sqrt{1+m_1^2}}+\frac{m_2x-y}{\sqrt{1+m_2^2}}]=0$$
$$or,\frac{(m_1x-y)^2}{1+m_1^2}-\frac{(m_2x-y)^2}{1+m_2^2}=0$$
$$or,\frac{(1+m_2^2)(m_1x-y)^2-(1+m_1^2)(m_2x-y)^2}{(1+m_1)^2(1+m_2)^2}=0$$
$$or,(1+m_2^2)(m_1^2x^2-2m_1xy+y^2)-(1+m_1^2)(m_2^2x^2-2m_2xy+y^2)=0$$
similarly,
simplifying above equation, we will get,
$$h(x^2-y^2)=(a-b)xy$$
which is the required equation of the bisectors of an angles between lines represented byax^{2}+2hxy+by^{2}=0.
proof:
Letax^{2}+2hxy+by^{2}+2gx+2fy+c=0 represents a pair of lines,
$$l_1x+m_1y+n_1=0.................(i)$$
$$l_2x+m_2y+n_2=0...................(ii)$$
The equation of line parallel to line (i) and passing through origin is given by,
$$l_1x+m_1y=0.................(iii)$$
and the equation of the line parallel to line (ii) and passing through origin is given by,
$$l_2x+m_2y=0.....................(iv)$$
Here,$$(l_1x+m_1y+n_1)(l_2x+m_2y+n_2=)=ax^2+2hxy+by^2+2gx+2fy+c$$
$$or,l_1l_2x^2+(l_1m_2+m_1l_2)xy+m_1m_2y^2+(l_1n_2+n_1l_2)x+(m_1n_2+m_2n_1)y+n_1n_2=ax^2+2hxy+by^2+2gx+2fy+c$$
Now,
comparing the coefficient of like terms on both sides, we get,
$$a=l_1l_2$$
$$2h=l_1m_2+m_1l_2$$
$$b=m_1m_2$$
Now, combining equation (iii) & (iv), we get,
$$(l_1x+m_1y)(l_2x+m_2y)=0$$
$$or,l_1l_2x^2+(l_1m_2+m_1l_2)xy+m_1m_2y^2=0$$
$$or,ax^2+2hxy+by^2=0$$
which is required...
solution:-
Given equation of a curve is,
$$ax^2+2hxy+by^2+2gx+2fy+c=0.......................................(i)$$
and given equation of a line is,
$$\frac{lx+my}{n}=1.........................................................(ii)$$
Now,
consider an equation,
$$ax^2+2hxy+by^2+2gx(\frac{lx+my}{n})+2fy(\frac{lx+my}{n})+c(\frac{lx+my}{n})^2=0.......................................(iii)$$
Since, here clearly we see that coordinates of P and Q satisfy equation (i) & (ii) and hence also satisfy equation (iii). so, locus of equation (iii) passes through P and Q where P and Q are the intersection points of acurve (i) and line (ii).
Equation (iii) being a homogeneous second-degree equation . so, it represents pair of lines through origin.
Thus, equation (iii) represents equation of two straight lines OP and OQ.
The homogeneous equation of second degree i.e, ax^{2}+2hxy+by^{2}=0 represents,
ASK ANY QUESTION ON Pair Of Straight Lines
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