# Intersection of Lines and Perpendicular Distance

Two straight lines intersect at a single point (x, y). The point satisfies the equations of both lines. We find this point by solving simultaneous equations.

Concurrent lines meet a single point. To show that lines are concurrent, solve two simultaneous equations to find the point of intersection. Then substitute this point of intersection into the third and subsequent lines to show that these lines also pass through the point.

### Examples:

1. Find the point of intersection between lines $2x-3y-3=0$ and $5x-2y-13=0$.

Solve simultaneous equations:

$\begin{matrix} 2x & - & 3y & - & 3 & = & 0 & \qquad \qquad (1) \\ 5x & - & 2y & - & 13 & = & 0 & \qquad \qquad (2) \end{matrix}\\ \begin{matrix} 4x & - & 6y & - & 6 & = & 0 & \qquad (1)\times 2 \\ 15x & - & 6y & - & 39 & = & 0 & \qquad (2)\times 3 \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \begin{matrix} -11x & \quad & \quad & + & 33 & = & 0 & \qquad (3)-(4) \end{matrix}\\ \qquad 11x\qquad =\qquad 33\\ \qquad \qquad \therefore \quad x=3\\ \\ \qquad Substitute\quad x=3\quad into\quad (1):\\ \qquad 2(3)-3y-3=0\\ \qquad 6-3y-3=0\\ \qquad 3y=3\\ \qquad \therefore \quad y=1\\ \\ So\quad the\quad point\quad of\quad intersection\quad is\quad (3,1).$

2. Show that the lines $3x-y+1=0$, $x+2y+12=0$ and $4x-3y-7=0$ are concurrent.

$\begin{matrix} 3x & - & y & + & 1 & = & 0 & \qquad (1) \\ x & + & 2y & + & 12 & = & 0 & \qquad (2) \\ 4x & - & 3y & - & 7 & = & 0 & \qquad (3) \\ 6x & - & 2y & + & 2 & = & 0 & \qquad (1)\times 2 \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \begin{matrix} 7x & \quad & \quad & + & 14 & = & 0 & \qquad (2)+(4) \end{matrix}\\ \qquad 7x\qquad =\quad -14\\ \qquad \qquad \therefore \quad x=-2\\ \\Substitute\quad x=-2\quad into\quad (1):\\ \qquad 3(-2)-y+1=0\\ \qquad -6-y+1=0\\ \qquad y=-5\\ \\ So\quad the\quad point\quad of\quad intersection\quad of\quad (1)\quad and\quad (2)\quad is\quad (-2,-5).\\ \\ Substitute\quad (-2,-5)\quad into\quad (3):\\ \qquad 4x-3y-7=0\\ \qquad 4(-2)-3(-5)-7=0\\ \\ \qquad LHS\quad =\quad RHS\\ \\ So\quad the\quad point\quad lies\quad on\quad line\quad (3)\\ \therefore \quad all\quad three\quad lines\quad are\quad concurrent.$

## Perpendicular Distance

Perpendicular distance is used to find the distance between a point and a line. If we look at the distance between a point and a line, there could be many distances.

So we choose the shortest distance, which is the perpendicular distance.

The perpendicular distance from $({ x }_{ 1 },{ y }_{ 1 })$ to the line $({ x }_{ 1 },{ y }_{ 1 })$ is given by:

$d=\cfrac { \left| a{ x }_{ 1 }+b{ y }_{ 1 }+c \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

### Examples:

1. Find the perpendicular distance of (4, -3) from the line $3x-4y-1=0$.

${ x }_{ 1 }=4,\quad { y }_{ 1 }=-3,\quad a=3,\quad b=-4,\quad c=-1\\ \\ d=\cfrac { \left| a{ x }_{ 1 }+b{ y }_{ 1 }+c \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \\ \\ d=\cfrac { \left| 3(4)+(-4)(-3)+(-1) \right| }{ \sqrt { { 3 }^{ 2 }+({ -4 })^{ 2 } } } \\ \\ d=\cfrac { \left| 12+12-1 \right| }{ \sqrt { 25 } } \\ \\ d=\cfrac { 23 }{ 5 }$

2. Prove that the line $6x+8y+20=0$ is a tangent to the circle ${ x }^{ 2 }+{ y }^{ 2 }=4$.

In this problem, we know the circle ${ x }^{ 2 }+{ y }^{ 2 }=4$ has the origin $(0,0)$ has a radius of $\sqrt { 4 } =2$ units, which should be the perpendicular distance if the statement above had been true. Therefore, we prove that the perpendicular distance from the line to the origin $(0, 0)$ is 2 units.
$d=\cfrac { \left| a{ x }_{ 1 }+b{ y }_{ 1 }+c \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \\ \\ d=\cfrac { \left| 6(0)+8(0)+20 \right| }{ \sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } } \\ \\ d=\cfrac { \left| 20 \right| }{ 100 } \\ \\ d=2\quad units\\ \\ \therefore \quad The\quad line\quad is\quad tangent\quad to\quad the\quad circle$