# Pair of Straight Lines

Let $A_1x+B_1y+C_1=0\text{ __(1)}$ and $A_2x+B_2y+C_2=0\text{ __(2)}$ be the equations of any two straight lines. Now, consider an equation $(A_1x+B_1y+C_1)(A_2x+B_2y+C_2)=0\text{ __(3)}$ The coordinates of any point lying in straight line $\text{(1)}$ will also satisfy $\text{(3)}$. Hence, all points on the straight line $\text{(1)}$ lie on $\text{(3)}$. Similarly, all points lying on the straight line $\text{(2)}$ lie on $\text{(3)}$. Conversely, the coordinates of any point lying on $\text{(3)}$ will satisfy $\text{(1)}$ and $\text{(2)}$.

Hence, equation $\text{(3)}$ represents a pair of the straight lines given by $\text{(1)}$ and $\text{(2)}$. If the left hand side of equation $\text{(3)}$ be expanded we get an equation of second degree in $x$ and $y$, i.e., an equation of the form $ax^2+2hxy+by^2+2gx+2fy+c=0$

Thus we see that the equation of a pair of straight lines is a second degree equation. But the converse is not always true. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.

# Homogeneous Equation

Homogeneous equation is an equation with two variables $x$ and $y$ in which each term has the same degree. If the degree of each term is $2$, then the equation is known as the homogeneous equation of degree two or a second degree homogeneous equation. The most general form of a homogeneous equation of degree two is $ax^2+2hxy+by^2=0$

## The Homogeneous Equation of Second Degree $ax^2+2hxy+by^2=0$, always Represents a Line Pair through the Origin.

The homogeneous equation of second degree in $x$ and $y$ is, $ax^2+2hxy+by^2=0$ If $b≠0$, then, the equation can be written as $\frac{a}{b}x^2+\frac{2h}{b}xy+y^2=0$ $\frac{a}{b}+\frac{2h}{b}\frac{y}{x}+\frac{y^2}{x^2}=0$ $\left(\frac{y}{x}\right)^2+\frac{2h}{b}\frac{y}{x}+\frac{a}{b}=0$ which is quadratic in $\frac{y}{x}$ having two roots $m_1$ and $m_2$. $\therefore\frac{y}{x}=m_1\text{ and }\frac{y}{x}=m_2$ $\text{or }y=m_1x\text{ and }y=m_2x$ These are two straight lines through origin.

If $b=0$, then the equation becomes, $ax^2+2hxy=0$ $x(ax+2hy)=0$ $\therefore x=0\text{ and }ax+2hy=0$ These equations again represent two straight lines through the origin. Hence, in all cases, the homogeneous equation will represent two straight lines through the origin.

Since $m_1$ and $m_2$ are two values of $\frac{y}{x}$ of $\left(\frac{y}{x}\right)^2+\frac{2h}{b}\frac{y}{x}+\frac{a}{b}=0$, we have, $m_1+m_2=-\frac{2h}{b}\text{ and }m_1m_2=\frac{a}{b}$

### Find a single equation representing the line pair $x+y+2=0$ and $x+2y+1=0$.

Given line pair is, $x+y+2=0$ $x+2y+1=0$ Combined equation is, $(x+y+2)(x+2y+1)=0$ $x^2+2xy+x+xy+2y^2+y+2x+4y+2=0$ $\therefore x^2+2y^2+3xy+3x+5y+2=0$

### Determine the lines represented by the equation $x^2+2xy+y^2-2x-2y-15=0$.

Given equation is, $x^2+2xy+y^2-2x-2y-15=0$ $x^2+(2y-2)x+(y^2-2y-15)=0$ which is quadratic in $x$. $\therefore x=\frac{-(2y-2)\pm\sqrt{(2y-2)^2-4(y^2-2y-15)}}{2}$ $x=-(y-1)\pm\sqrt{y^2-2y+1-y^2+2y+15}$ $x=-(y-1)\pm 4$ $\therefore x=-y+1+4\text{ and }x=-y+1-4$ $x+y-5=0\text{ and }x+y+3=0$

### Find the equation of the straight lines through the origin and at right angles to the lines $x^2-5xy+4y^2=0$.

Given line pair is, $x^2-5xy+4y^2=0$ Let $y=m_1x$ and $y=m_2x$ be the two lines represented by the above line pair. Then, $m_1+m_2=-\frac{2h}{b}=\frac{5}{4}$ $m_1m_2=\frac{a}{b}=\frac{1}{4}$ Since required lines are perpendicular to the given lines, so, slope of the required lines are $-\frac{1}{m_1}$ and $-\frac{1}{m_2}$. Also, the lines pass through the origin. Thus, equations of required lines are, $y-0=-\frac{1}{m_1}(x-0)\text{ and }y-0=-\frac{1}{m_2}(x-0)$ $x+m_1y=0\text{ and }x+m_2y=0$ Hence, combined equation is, $(x+m_1y)(x+m_2y)=0$ $x^2+(m_1+m_2)xy+m_1m_2y^2=0$ $x^2+\frac{5}{4}xy+\frac{1}{4}y^2=0$ $\therefore 4x^2+5xy+y^2=0$

### Find the equation of the straight lines passing through $(1,1)$ and parallel to the lines represented by $x^2-5xy+4y^2+x+2y-2=0$.

Given line pair is, $x^2-5xy+4y^2+x+2y-2=0$ Taking homogeneous part only, $x^2-5xy+4y^2=0$ Let $y=m_1x$ and $y=m_2x$ be the two lines represented by the above line pair. Then, $m_1+m_2=-\frac{2h}{b}=\frac{5}{4}$ $m_1m_2=\frac{a}{b}=\frac{1}{4}$ Since required lines are parallel to the given lines, so, slope of the required lines are $m_1$ and $m_2$. Also, the lines pass through $(1,1)$. Thus, equations of required lines are, $y-1=-\frac{1}{m_1}(x-1)\text{ and }y-1=-\frac{1}{m_2}(x-1)$ $y-1-m_1(x-1)=0\text{ and }y-1-m_2(x-1)=0$ Hence, combined equation is, $[(y-1)-m_1(x-1)][(y-1)-m_2(x-1)]=0$ $(y-1)^2-(m_1+m_2)(x-1)(y-1)+m_1m_2(x-1)^2=0$ $(y-1)^2-\frac{5}{4}(x-1)(y-1)+\frac{1}{4}(x-1)^2=0$ $4y^2-8y+4-5xy+5x+5y-5+x^2-2x+1=0$ $\therefore x^2-5xy+4y^2+3x-3y=0$

### Find the condition that one of the lines $ax^2+2hxy+by^2$ may coincide with ine of the lines $a_1x^2+2h_1xy+b_1y^2=0$.

Given pair of straight lines are, $ax^2+2hxy+by^2=0\text{ __(1)}$ $a_1x^2+2h_1xy+b_1y^2=0\text{ __(2)}$ Let $y=mx\text{ __(3)}$ of pair $\text{(1)}$ coincide with $y=m’x\text{ __(4)}$ of pair $\text{(2)}$. Then, $m’=m$.

Eliminating $y$ between $\text{(1)}$ and $\text{(3)}$, $ax^2+2hx.mx+b.m^2x^2=0$ $bm^2+2hm+a=0\text{ __(A)}$ Eliminating $y$ between $\text{(2)}$ and $\text{(4)}$, $a_1x^2+2h_1x.m’x+b_1.m’^2x^2=0$ $a_1+2h_1m’+b_1m’^2=0$ $b_1m’^2+2h_1m’+a_1=0$ $\therefore b_1m^2+2h_1m+a_1=0$ Solving $\text{(A)}$ and $\text{(B)}$ for $m^2$ and $m$, $\frac{m^2}{2ha_1-2h_1a}=\frac{m}{ab_1-a_1b}=\frac{1}{2h_1b-2hb_1}$ $\therefore m=\frac{2(ha_1-h_1a)}{ab_1-a_1b}\text{ and }m=\frac{ab_1-a_1b}{2(h_1b-hb_1)}$ $\therefore\frac{2(ha_1-h_1a)}{ab_1-a_1b}=\frac{ab_1-a_1b}{2(h_1b-hb_1)}$ $4(ha_1-h_1a)(h_1b-hb_1)=(ab_1-a_1b)^2$ This is the required condition.