How To Find Where Two Lines Intersect With Equations

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

Intersection of Lines

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

To discover the intersection of ii lines, you offset need the equation for each line. At the intersection, $10$ and $y$ accept the same value for each equation. This means that the equations are equal to each other. We can therefore solve for $10$ . Substitute the value of $x$ in one of the equations (it does not matter which) and solve for $y$ .

Detect the intersection of the lines $y = 3x - 3$ and $y = ii.3x + 4$ .

Nosotros have

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

Thus, the intersection betoken is $(10, 27)$ . $_\foursquare$

Angle between the lines:

Ange between the lines is given by $\tan(\theta )=\frac { { m }_{ ane }-{ g }_{ 2 } }{ one+{ yard }_{ ane }{ m }_{ 2 } } ,$ where ${m}_{one}$ is the slope of the kickoff line, ${one thousand}_{2}$ is the slope of the second line, and $\theta$ is the angle betwixt them.

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

Second-caste equation representing a pair of straight lines:

Homogeneous equations (theorem):
A second-caste homogeneous equation in $10$ and $y$ always represents a pair of straight lines (existent or imaginary) passing through the origin.

If $h^2 \geq ab$ , the equation $ax^2+2hxy+by^two=0$ 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 caste $1$ ) of the course $y=mx$ and $y=nx$ .

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

Consider the curve $10^two -2y^two +axy+3y-1=0$ .

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

Comparing the above equation with the full general one and substituting in the $2$ atmospheric condition, we find that $\begin{aligned} i(-ii)(-1) +iii(0)(a) - one\frac{9}{iv}- two(0) - (-1)\frac{a^2}{4}&=0\\ ii-\frac{9}{4} + \frac{a^ii}{4}&=0\\ a^two &= i. \end{aligned}$ Checking if $h^2>ab$ or $\frac{a^2}{two} > (-2)$ gives $\frac{ane}{four} >(-2).$

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

Let united states of america detect the equation of the direct lines joining the origin and the points of intersection of the curve

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

and the line

$60+my+n=0.$

This can be rewritten every bit

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

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 $\frac{lx+my}{-n}=one$ , nosotros write the pair of lines as

$\begin{aligned} ax^2+2hxy+by^ii+\left(2gx+2fy\right)\left(one\right)+c\left(1\correct)^2&=0\\ ax^2+2hxy+by^two+\frac{\left(2gx+2fy\correct)\left(lx+my\right)}{\left(-north\right)}+c\left(\frac{\left(lx+my\correct)}{\left(-northward\right)}\correct)^ii&=0. \finish{aligned}$