Intersection of Lines

Lines that are non-coincident and non-parallel intersect at a unique point. Lines are said to intersect each other if they cut each other at a point. By Euclid's lemma two lines can have at most $1$ point of intersection. In the figure below lines $L1$ and $L2$ intersect each other at point $P.$ Three or more lines when met at a single point are said to be concurrent and the point of intersection is point of concurrency.

Intersection of Lines

In the figure above, point $P=(p, q)$ satisfies both equations.

To find the intersection of two lines, you first need the equation for each line. At the intersection, $x$ and $y$ have the same value for each equation. This means that the equations are equal to each other. We can therefore solve for $x$. Substitute the value of $x$ in one of the equations (it does not matter which) and solve for $y$.

Find the intersection of the lines $y = 3x - 3$ and $y = 2.3x + 4$.

---

We have

$$\begin{aligned} 3x - 3 &= 2.3x + 4\\ 3x - 2.3x &= 4 + 3\\ 0.7x &= 7\\ \Rightarrow x &= 10\\ \Rightarrow y &= 3(10) - 3\\ &= 27. \end{aligned}$$

Thus, the intersection point is $(10, 27)$. $_\square$

Angle between the lines:

Ange between the lines is given by $$\tan(\theta )=\frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } ,$$ where ${m}_{1}$ is the slope of the first line, ${m}_{2}$ is the slope of the second line, and $\theta$ is the angle between them.

For two lines intersecting at right angle, ${ m }_{ 1 }{ m }_{ 2 } =-1.$

$$$$ Second-degree equation representing a pair of straight lines:

Homogeneous equations (theorem):
A second-degree homogeneous equation in $x$ and $y$ always represents a pair of straight lines (real or imaginary) passing through the origin.

If $h^2 \geq ab$, the equation $$ax^2+2hxy+by^2=0$$ represents a pair of straight lines passing through the origin. This equation can be considered a quadratic in $y$ and can be solved to obtain two equations (of degree $1$) of the form $y=mx$ and $y=nx$.

However, general equation in degree $2$ $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ will represent a pair of straight lines if and only if $$abc+ 2fgh- af^2 -bg^2 -ch^2 =0\quad \text{ and }\quad h^2 - ab > 0.$$ The angle $\theta$ between these lines satisfies $$\tan \theta=\frac{\sqrt{h^2-ab}}{|a-b|}.$$

Consider the curve $x^2 -2y^2 +axy+3y-1=0$.

Find the values of $a$ for which this equation represents a pair of straight lines.

---

Comparing the above equation with the general one and substituting in the $2$ conditions, we find that $$\begin{aligned} 1(-2)(-1) +3(0)(a) - 1\frac{9}{4}- 2(0) - (-1)\frac{a^2}{4}&=0\\ 2-\frac{9}{4} + \frac{a^2}{4}&=0\\ a^2 &= 1. \end{aligned}$$ Checking if $h^2>ab$ or $\frac{a^2}{2} > (-2)$ gives $\frac{1}{4} >(-2).$

Therefore, $a=1$ or $a = -1. \ _\square$

Let us find the equation of the straight lines joining the origin and the points of intersection of the curve

$$ax^2+2hxy+by^2+2gx+2fy+c=0$$

and the line

$$lx+my+n=0.$$

This can be rewritten as

$$\begin{aligned} lx+my&=-n\\ \frac{lx+my}{-n}&=1. \end{aligned}$$

Let $A$ and $B$ be the points of intersection of the curve and the line. In order to make the pair of lines homogeneous with the help of $\frac{lx+my}{-n}=1$, we write the pair of lines as

$$\begin{aligned} ax^2+2hxy+by^2+\left(2gx+2fy\right)\left(1\right)+c\left(1\right)^2&=0\\ ax^2+2hxy+by^2+\frac{\left(2gx+2fy\right)\left(lx+my\right)}{\left(-n\right)}+c\left(\frac{\left(lx+my\right)}{\left(-n\right)}\right)^2&=0. \end{aligned}$$

So, this is locus through points $A$ and $B$. Also it represents the homogeneous equation of second degree in $x$ and $y$ through origin.