Notes on Pair Of Straight Lines | Grade 11 > Mathematics > Coordinate Geometry straight line | KULLABS.COM

• Note
• Things to remember

### Introduction

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 ax2+2hxy+by2=0, always represents a pair of straight lines through the origin.

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 bx2 ,

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 m1 and m2 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.

#### Angle between the line pair represented by ax2+2hxy+by2=0

Solution,

Let two lines represented by ax2+2hxy+by2=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 byax2+2hxy+by2=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, ax2+2hxy+by2=0 represents,

• two real and distinct lines if h2-ab>0
• two real and coincident lines if h2-ab=0
• two imaginary lines if h2-ab <0

#### Condition that the general equation of second degree may represent a line pair

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

h2y2+2ghy+g2-aby2-2afy-ca is a perfect square.

i.e. (h2-ab)y2+(2gh-2af)y+(g2-ca) is a perfect square.

Its discriminant i.e. (B2-4AC) 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.

#### Equation of the bisectors of the angle between the pair of lines represented by ax2+2hxy+by2=0

The two lines represented by ax2+2hxy+by2=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 byax2+2hxy+by2=0.

#### If the equations ax2+2hxy+by2+2gx+2fy+c=0 represent a pair of lines, then ax2+2hxy+by2=0 represent a pair of lines through the origin parallel to the above pair.

proof:

Letax2+2hxy+by2+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...

#### If the equation ax2+2hxy+by2+2gx+2fy+c=0 represents a curve. Then find the equation of line joining origin to the point of intersection of this curve and the line lx+my=n

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, ax2+2hxy+by2=0 represents,

• two real and distinct lines if h2-ab>0
• two real and coincident lines if h2-ab=0
• two imaginary lines if h2-ab <0
• equation of the bisectors of an angles between lines represented byax2+2hxy+by2=0 is $$h(x^2-y^2)=(a-b)xy$$
.