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

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

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 \[\begin{align} 1(-2)(-1) +3(0)(a) - 1\frac{9}{4}- 2(0) - (-1)\frac{a^2}{4}&=0\\ 2-\frac{9}{4} + \frac{a^2}{4}&=0\\ a^2 &= 1. \end{align}\] 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

\[\begin{align} lx+my&=-n\\ \frac{lx+my}{-n}&=1. \end{align}\]

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

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

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