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# Straight Line

Analysis of the Straight line and the Area of a Triangle

## Overview

By using Pythagoras, the distance between the points <strong>A</strong> and<strong>B</strong> can be found from:
$\sqrt{(x_1&space;-&space;x_2)^2&space;+&space;(y_1&space;-&space;y_2)^2}$

And the gradient of the line A B is given by:
$\frac{(y_2&space;-&space;y_1)}{(x_2&space;-&space;x_1)}$

## Coordinates Of A Point Dividing A Line Ab

Suppose the coordinates of the required point P are $\inline&space;(X,Y)$ that divide a line AB in the ratio of $\inline&space;(\lambda:\mu)$.

Then by parallels:
$\frac{x_2&space;-&space;X}{X&space;-&space;x_1}&space;=&space;\frac{\mu&space;}{\lambda&space;}$
and so
$X(\lambda&space;&space;+&space;\mu&space;)&space;=&space;\lambda&space;\;x_2&space;+&space;\mu&space;\;x_1$

Similarly for the y coordinate and so the coordinates of P are:
$\frac{\lambda&space;\;x_2&space;+&space;\mu&space;\;x_1}{\lambda&space;&space;+&space;\mu&space;}$
and
$\frac{\lambda&space;\;y_2&space;+&space;\mu&space;\;y_1}{\lambda&space;&space;+&space;\mu&space;}$

N.B. If P is not between A and B, the ratio AP/PB is negative and the same formula holds good, provided that $\inline&space;\mu$ is taken to be negative. As a particular example. putting the ratio as 1 gives the coordinates of the mid point of the line as:
$\frac{1}{2}(x_1&space;+&space;x_2)$
and
$\frac{1}{2}(y_1&space;+&space;y_2)$

## Centre Of Gravity For A Triangle A B C

If A' is the mid point of B C

$A'\;\equiv&space;\;\left[\frac{1}{2}(x_3&space;-&space;x_2)\;and\;\frac{1}{2}(y_3&space;-&space;y_2)&space;\right]$
But G divides AA' in the ratio of 2 : 1. The coordinates of G are therefore :
$\left(&space;2\left(\frac{x_3&space;+&space;x_2}{2}&space;\right)&space;+&space;x_1&space;\right)\div&space;3&space;=&space;\frac{1}{3}\left(x_1&space;+&space;x_2&space;+&space;x_1&space;\right)$

The y coordinates follow in a similar manner and so <strong>Each coordinate of the centre of Gravity of a triangle is one third of the sum of the coordinates of the vertices.</strong>

## The Area Of The Triangle A B C

From the diagram it can be seen that:
$ABC\;&space;=&space;\;Trapezium&space;\;APRC&space;&space;+&space;&space;Trapezium\;&space;CRQB\;&space;-&space;&space;Trapezium\;&space;APQB$
$=&space;\frac{1}{2}(y_3&space;+&space;y_1)(x_3&space;-&space;x_1)&space;+&space;\frac{1}{2}(y_2&space;+&space;y_3)(x_2&space;-&space;x_3)&space;-&space;\frac{1}{2}(y_1\;+y_2)(x_2&space;-&space;x_1)$
$=&space;\frac{1}{2}\left(x_1y_2&space;-&space;x_2y_1&space;+&space;x_2y_3&space;-&space;x_3y_2&space;+&space;x_3y_1&space;-&space;x_1y_3&space;\right)$
N.B. It often helps to use the particular form of this equation when C becomes the Origin. The area of this triangle is then given by:
$=&space;\frac{1}{2}\left(x_1y_2&space;-&space;x_2y_1&space;\right)\;\;\;\;\;\;\;\;i.e.\;(x_3&space;=&space;y_3&space;=&space;0)$

## The Equation Of A Straight Line

If (x y) is a point on the line joining $\inline&space;(x_1&space;\,y_1\;and\;x_2\,y_2)$, the area of the triangle formed by the three points is zero.

$\therefore\;\;\;\;\;\;x_1y_2\;-x_2y_1&space;+&space;x_2y&space;-&space;xy_2&space;+&space;xy_1&space;-&space;x_1y&space;=&space;0$

$or\;\;\;\;\;\;x_2(y&space;-&space;y_1)&space;-&space;y_2(x&space;-&space;x_1)&space;=&space;yx_1&space;-&space;xy_1$

$=&space;x_1(y&space;-&space;y_1)&space;-&space;y_1(x&space;-&space;x_1)$

$\therefore\;\;\;\;\;(x_2&space;-&space;x_1)(y&space;-&space;y_1)&space;=&space;(y_2&space;-&space;y_1)(x&space;-&space;x_1)$

$\;or\;\;\;\;\;\;[\frac{y&space;-&space;y_1}{x&space;-&space;x_1}&space;=&space;\frac{y_2&space;-&space;y_1}{x_2&space;-&space;x_1}$

This is therefore the equation of a straight line joining the points A and B and since it is of the first degree in x and y, it also shows that any straight line must be represented by an equation in the first degree.

Since $\inline&space;\frac{y_2&space;-&space;y_1}{x_2&space;-&space;x_1}$ is the gradient of the line , it's equation may be written as:-

$\frac{y&space;-&space;y_1}{x&space;-&space;x_1}&space;=&space;m$

$\theta&space;&space;=&space;\beta&space;&space;-&space;\alpha$

Where m is the gradient of the line.

Intercept Form

To find the equation of the line which makes intercepts a and b on the axes. We want the line joining (a 0) to (b 0). So the equation is:-

$\frac{y&space;-&space;0}{x&space;-&space;a}&space;=&space;\frac{b&space;-&space;0}{0&space;-&space;a}$

$i.e.\;\;\;\;\;\;-a\,y&space;=&space;b\,x&space;-&space;b\,a$

$\therefore\;\;\;\;\;\;\frac{x}{a}&space;+&space;\frac{y}{b}&space;=&space;1$

To find the equation of a straight line of gradient m which makes an intercept c on the y axis.

The intercepts are obviously c and - c/m and so the equation is given by:-
$-&space;\frac{x}{c/m}&space;+&space;\frac{y}{c}&space;=&space;1$

$\therefore\;\;\;\;\;y&space;=&space;m\,x&space;+&space;c$

The Polar Form

Yo find the equation of a straight line such that the from the origin is of length p and makes an angle $\inline&space;\alpha$ with the x-axis.

If (x,y) are the coordinates of any point on the line. From the diagram we can see that :-

$ON&space;=&space;OS&space;+&space;PR$

$\therefore\;\;\;\;\;\;p&space;=&space;x\;cos\,\alpha&space;&space;+&space;y\;sin\,\alpha$

This then is the equation required.

## The Angle Between Two Lines.

To find the angle between to lines of gradient m and t.

From the Diagram it can be seen that:-

$\theta&space;&space;=&space;\beta&space;&space;-&space;\alpha$

$\therefore\;\;\;\;\;tan\,\theta&space;&space;=&space;tan\,(\beta&space;&space;-&space;\alpha)$

$=&space;\frac{tan\;\beta&space;&space;-&space;tan\;\alpha&space;}{1&space;-&space;tan\;\beta&space;\;tan\;\alpha&space;}$

$But\;since\;\;\;\;\;\;\;tan\;\beta&space;&space;=&space;m\;\;\;and\;\;\;tan\;\alpha&space;&space;=&space;t$

$tan\;\theta&space;&space;=&space;\frac{m&space;-&space;t}{1&space;+&space;mt}$

If the lines are parallel, since tan 0 is zero, m = t

and if the lines are perpendicular, since tan 90 is infinite, mt = - 1

i.e. The product of the gradients of perpendicular lines is - 1.

## Example.

Write down the equation of the line through (1,2)which are parallel and perpendicular to
$3x\;&space;-&space;&space;4y\;&space;=&space;&space;7$

For the parallel line keep the x and y terms unaltered. The equation required is 3x - 4y - c = 0 and since the line passes through (1,2) , the value of C can be found by substituting x = 1 and y = 2 in the equation. The parallel line is thus :-

$3x&space;-&space;4y\;=&space;-&space;5$

For the perpendicular line, interchange the coefficients of x and y and alter the sign between them. The line becomes 4x + 3y = K and as before the value of the constant is found by substituting x = 1 and y = 2. The equation of the perpendicular line is therefore:-

$4x&space;+&space;3y&space;=&space;10$

## The Length Of The Perpendicular

To find the length of the perpendicular from (x', y') to the line ax + by + c = 0

Suppose that the perpendicular makes an angle$\inline&space;\alpha$ with the x axis. If the length of the perpendicular is p then the coordinates of it's foot are:-

$x'&space;+&space;p\,cos\,\alpha&space;\;\;\;and\;\;\;y'\;+p\,sin\,\alpha$

This point lies on the line ax + by + c = 0 and therefore:-

$a(x'&space;+&space;p\,cos\,\alpha)&space;&space;+&space;b(y'&space;+&space;p\,sin\,\alpha&space;)&space;+&space;c&space;=&space;0$

$or\;\;\;\;\;p(a\;cos\,\alpha&space;&space;+&space;b\;sin\,\alpha&space;)\;=&space;-&space;(ax'&space;+&space;by'&space;+&space;c)$

But since the product of perpendicular lines is - 1

$tan\;\alpha&space;\left(-&space;\frac{a}{b}&space;\right)\;=&space;-&space;1$

$and\;\therefore\;\;\;\;\;\;tan\;\alpha&space;&space;=&space;\frac{b}{a}$

$\therefore\;\;\;\;\;a\;cos\,\alpha&space;&space;+&space;b\;sin\,\alpha&space;&space;=&space;a\times&space;\frac{a}{\sqrt{a^2&space;+&space;b^2}}&space;+&space;b\times&space;\frac{b}{\sqrt{a^2&space;+&space;b^2}}&space;=&space;\sqrt{a^2&space;+&space;b^2}$

$p\;=&space;-&space;\frac{ax'&space;+&space;by'&space;+&space;c}{\sqrt{a^2&space;+&space;b^2}}$

The minus sign is of no great significance in itself ( Since we have a square root in th denominator) but the comparison between the signs of the perpendicular is of the utmost importance. If these perpendiculars are of the same sign, the points are on the same side of the line. If they are of different signs the points are on opposite sides. The square root of the denominator is assumed to have it's positive value throughout and so will not affect the comparison. Hence all we need to do is to substitute the points in the lines themselves.

Example. Are the points (i.2) and (3,1) on the same side or on opposite sides of the line 3x - 4y - 1 = 0.

If x = 1. y = 2 then the value of 3x - 4y - 1 is 3 - 8 - 1 i.e.-ve

If x = 3, y = 1 then the value of 3x - 4y - 1 is 9 - 4 - 1 i.e.+ve

So the points are on either side of the line.

## Angle Bisectors.

To find the equation of the angle bisectors between the lines ax + by + c = 0 and Ax + By + C = 0

Use the geometrical property that the perpendiculars from any point on either angle bisector to the two lines are equal.

$\therefore\;\;\;\;\;\;\;\frac{ax&space;+&space;by&space;+&space;c}{\sqrt{a^2&space;+&space;b^2}}&space;=&space;\pm&space;\;\frac{Ax&space;+&space;By&space;+&space;C}{\sqrt{A^2&space;+&space;B^2}}$

These are the required pair of lines.

It is sometimes necessary to distinguish which of these is the internal and which is the external bisector and a method of doing this is shown in the following example.

Example

Find the incentre of the triangle formed by the following three lines:-

x + 2y - 10 = 0; 2x + y - 9 = 0 and x - 2y - 2 = 0

It is helpful to draw a diagram showing the relative positions of the lines. If (x,y) is the incentre the length of the perpendicular from (x,y) to the line 2x + y - 9 = 0 is given by:-

$\frac{2x&space;+&space;y\;-9}{\sqrt{5}}$

If the coordinates of the origin are substituted into this , the result is a negative quantity but (x,) and the origin are on opposite sides of the line and so:-

$\frac{2x&space;+&space;y\;-9}{\sqrt{5}}\;\;\;\;is\;positive$

The perpendicular from (x.y) to the line x - 2y - 2 = 0 is given by:-

$\frac{x&space;-&space;2y&space;-&space;2}{\sqrt{5}}$

The origin substituted in this will give a negative expression and as (x,y) and the origin are on the same side.

$\frac{x&space;-&space;2y&space;-&space;2}{\sqrt{5}}\;\;\;\;is\;&space;negative$

The perpendicular from (x,y) to the line x + 2y - 10 = 0 is given by:-
$\frac{x&space;+&space;2y&space;-&space;10}{\sqrt{5}}$

The origin substituted in this expression gives a negative quantity and since (x,y) and the origin are on the same side :-
$\frac{x&space;+&space;2y&space;-&space;10}{\sqrt{5}}\;\;\;\;\;is\;negative$

$\therefore\;\;\;\;\;\frac{2x&space;+&space;y&space;-&space;9}{\sqrt{5}}&space;&space;=&space;\frac{x&space;-&space;2y&space;-&space;2}{\sqrt{5}}&space;\;=\&space;;\frac{x&space;+&space;2y&space;-&space;10}{\sqrt{5}}$

From which$\inline&space;3x&space;-&space;y&space;=&space;11\;\;\;\;and\;\;\;\;3x&space;+&space;3y&space;=&space;19$

Solving these two equations gives the coordinates of the incentre as (4.5, 2)

Taking the alternative signs in the equations will give the ex-centres

## A Line Through The Intersection Of Two Given Lines

If l = 0 and l' = 0 are the equations of any two straight lines , then $\inline&space;l&space;+&space;\lambda&space;l'&space;=&space;0$ will represent e line passing through their point of intersection for all values of $\inline&space;\lambda$.

Since l and l' are expressions of the first degree so must be $\inline&space;l&space;+&space;\lambda&space;l'$ and therefore$\inline&space;l&space;+&space;\lambda&space;l'&space;=&space;0$ must be a straight line. The coordinates of the point of intersection of l and l' will make both l and l' equal zero and this will make l + l' also = 0. Therefore the line l = l' = 0 passes through the point of intersection og l and l'.

This is of particular use in finding the equation of the line which joins the point of intersection of two given lines to the origin.

Example.

Find the equation of the line joining the point of intersection of

2x + y - 3 = 0 and x + 3y + 8 = 0

To the origin.

Choose multiples of the lines so that on addition the constant terms will vanish. The resulting equation is:-

$8(2x&space;+&space;y&space;-&space;3)&space;+&space;3(x&space;+&space;3y&space;+&space;8)&space;=&space;0$

$\therefore\;\;\;\;\;\;19x&space;+&space;17y&space;=&space;0$

This represents a straight line through the the intersection of the two lines and it certainly passes through the origin.

### Example 1

Wr1te down the equation of the line connecting the points (1,3) and (-2,4)

$Substituting\;values\;\;\;\frac{y&space;-&space;3}{x&space;-&space;1}&space;=&space;\frac{4&space;-&space;3}{-2&space;-&space;1}$

$\therefore\;\;\;\;\;-3y&space;+&space;9&space;=&space;x&space;-&space;1$

$Thus\;\;\;\;\;\;3y&space;+&space;x&space;=&space;10$

### Example 2

Write down the equation of the line which passes through (1,-1) and has a gradient of 2

Substituting in the coordinates and m = -3

The equation becomes y = 2x - 3

### Example 3

Find the equation of the line such as it's perpendicular to the origin is of length 3 and makes an angle of 30 degrees with the x axis.

From the graph it can be seen that the required line passes through $\inline&space;\frac{3\sqrt{3}}{2}\;and\;\frac{3}{2}$. It can also be seen that the gradient of the line is - tan 60

Substituting in y = mx + c

$\frac{3}{2}\/=\/-&space;\sqrt{3}\times&space;\frac{3\sqrt{3}}{2}&space;+&space;c$

From Which c = 6

$\therefore&space;\;\;\;\;\;\;y\;+\sqrt{3}\;x&space;=&space;6$

### Example 4

Find the line which passes through (1,2) and is parallel to 3x + 4y = -7

Therefore substituting the coordinates in the given equation
$3\times&space;1&space;+&space;4\times&space;2&space;=&space;c\;=&space;-&space;11$
.

$\therefore\;\;\;\;\;3x&space;+&space;4y\;=&space;-&space;11$

### Example 5

find the equation of the line which passes through (1, -7) and is parallel to x = 2y = 5

$y\;=&space;-&space;\frac{1}{2}\;x&space;-&space;\frac{5}{2}\;\;\;\;\;\;therefore\;\;&space;m&space;=&space;-\frac{1}{2}\;\;\;and\;\;&space;t&space;=&space;2$

$\therefore\;\;\;\;\;the\;new\;line\;is\;y&space;=&space;2x&space;+&space;K$

Substituting in the coordinates through which the line must pass

$-&space;7&space;=&space;2&space;+&space;K\;\;\;\;\;\;or\;\;\;\;\;k\;=&space;-&space;9$

Thus the equation of the new line is y = 2x - 9

### Example 6

Find the area of the triangle formed by the points (1,1) (2,2) (3,5)

From the question $\inline&space;x_1&space;=&space;1\;\;x_2&space;=&space;2\;\;x_3&space;=&space;3\;\;\;y_1&space;=&space;1\;\;y_2&space;=&space;2\;\;y_3&space;=&space;5$

Substituting in the equation:-

$Area\;of\;the\;triangle&space;=&space;\frac{1}{2}(1&space;-&space;2&space;+&space;10&space;-&space;6&space;+&space;2&space;-&space;5)&space;=&space;1$