`(a)` `text(Distance formula : )`

=>The distance between two points `P (z_1)` and `Q (z_2 )` is given by : `P Q = |z_2- z_1| `

=>The distance of a point z from origin, `|z_0|=|z|`

=>Three points A `(z_1)`, B` (z_2)` and C `(z_3 )` are collinear, then `AB + BC = AC`

`| z_1 - z_2| + | z_2 - z_3 | = | z_1 - z_3 | `


`(b)text (Equation of the Perpendicular Bisector : )`

If P `(z_1)` and Q `(z_2)` are two fixed points and R (z) is moving point, such that it is always at equal distance from P `(z_1)` and Q `(z_2)`

`PR=QR`

`| z - z_1| = | z - z_2|`

`z (barz_1 - barz_2 ) + barz (z_1 - z_2) = z_1barz_1 - z_2 barz_2`


`z (barz_1 - barz_2 ) + barz (z_1 - z_2) = |z_1|^2 - |z_2|^2`

Hence, z lies on the perpendicular bisectors of `z_1` and `z_2.`


`(c) text(Section Formula :)`

If R (z) divides the joining of `P (z_1)` and `Q (z_2)` in the ratio `m_1 : m_2 \ \ \ \ (m_1, m_2 > 0).`

(i) If R (z) divides the segment PQ internally in the ratio of `m_1 : m_2,` then

`z = ((m_1z_2) + (m_2z_1))/(m_1+m_2)`

(ii) If R (z) divides the segment PQ externally in the ratio of `m_1 : m_2 ,` then

`z = ((m_1z_2) - (m_2z_1))/(m_1-m_2)`


`(d)text( Area of Triangle : )`

If `z_1 , z_2` and `z_3` are the affixes of the vertices of a triangle, then its area

`1/4 |(z_1,barz_1,1),(z_2,barz_2,1),(z_3,barz_3,1)|`



`(e)text (Equation of a Straight line :)`

(i) Parametric form Equation of the straight line joining the points having affixes `z_1` and `z_2` is

`z = tz_1 + (1- t) z_2 ,` where ` t in R ~ { 0} `

(ii) Non-parametric form Equation of the straight line joining the points having affixes `z_1` and `z_2` is

`|(z,barz,1),(z_1,barz_1,1),(z_2,barz_2,1)| = 0`

`z (barz_1 - barz_2 ) - barz (z_1 - z_2 ) + z_1 barz_2 - barz_1 z_2 = 0`

`text(Funda : )`

If `z_1, z_2` and `z_3` are collinear, then `|(z,barz,1),(z_1,barz_1,1),(z_2,barz_2,1)| = 0`

or `sumbarz_1(z_2 - z_3) = 0`

(iii) General form The general equation of a straight line is of the form `bara z +a barz + b = 0,` where a is a complex number and b is a real number.

(iv) Slope of the line `bara z +a barz + b = 0 `

`text(Remember :)`

If `alpha_1` and `alpha_2` are the complex slopes of two lines on the argand plane, then prove that the lines are

(i) perpendicular, if `alpha_1 + alpha_2 = 0`

(ii) parallel, if `alpha_1 = alpha_2`

(iii)The equation of a line parallel to the line `bara z + abarz + b = 0` is `bara z + a barz + lambda = 0,`

(v) Length of perpendicular from a given line `bara z + abarz + b = 0` is `bara z - abarz + ilambda = 0`

`(e)text( Length of perpendicular from a given point on a given lilne :)`

The length of perpendicular from a point `P (z_1)` to the line

`bara z + a barz + b = 0` is given by `|baraz_1 + a barz_1 + b I /(2a)`


`(f) text(Circle : ) `

The equation of a circle whose centre is at point affix `z_0` and radius r, is `| z- z_0 |= r .`

If the centre of the circle is at origin and radius r, then its equation is `|z| = 0`

`=>` (i) General Equation of a Circle The general equation of the circle is `zbarz+baraz+abarz+b=0,`

`text(Rule to find the centre and radius of a circle :)`

Make the coefficient of `zbarz` equal to 1 and right hand side equal to zero.

1. The centre of circle will be `= (-a) = (-` coefficient of `barZ).`

2. Radius = `sqrt(|a|^2 - text(constant term))`

`(ii)` `text(Equation of Circle Through Three Non-Collinear Points)`

Let `A (z_1), B(z_2 ), C (z_3 )` be three points on the circle and P (z) be any point on
the circle, then

`angle ACB = angle APB`

Using Coni method,

In `triangle ACB , (z_2-z_3)/(z_1-z_3) = (BC)/(CA) e^(i theta)..................(i)`

In `triangle APB , (z_2-z)/(z_1-z) = (BP)/(CA) e^(i theta)....................(ii)`

From Eqs. (i) and (ii), we get

`((z-z_1)(z_2-z_3))/((z-z_2)(z_1-z_3)) = text(Real)......................(iii)`

`text(Remember)`

If four points `z_1 , z_2 , z_3 , z_4` are concyclic, then `((z_4-z_1)(z_2-z_3))/((z_4-z_2)(z_1-z_3)) =` real [replacing z
by `z_4` in Eq. (iii)]

or `arg[((z_2-z_3)(z_4-z_1))/((z_1-z_3)(z_4-z_2))] = pi,0`


`(iii)` `text(Equation of Circle in Diametric Form)`

If end points of diameter represented by `A (z_1)` and `B(z_2)` and `P (z)` is any
point on circle.

` therefore angleAPB=90^o`

`therefore` Complex slope of PA + Complex slope of PB = 0

`=>((z-z_1)/(barz-barz_1)) + ((z-z_2)/(barz-barz_2)) = 0`

Hence, `(z-z_1)(barz- barz_2) + (z- z_2)(barz- barz_1) =0`

which is required equation of circle in diametric form.




`(g) text(Equation of Parabola :)`

Now, for parabola

SP=PM

`|z-a| = |z- barz + 2a|/2`

or `z barz - 4a(z+barz) = 1/2 {z^2 + (barz)^2}`

where `a in R` (focus) , directrix is `z + barz + 2a =0`


`(h)` `text(Equation of Ellipse)`

For ellipse

`SP +S' P== 2a`

`=> |z - z_1| + |z-z_2| = 2a`

when , `2a > |z_1 - z_2|` [since, eccentricity< 1]

Then, point z describes a ellipse having foci at `z_1` and `z_2` and `a in R^(+)`


`(i)` `text(Equation of Hyperbola)`

For hyperbola

`=> |z - z_1| - |z-z_2| = 2a`

when , `2a > |z_1 - z_2|` [since, eccentricity > 1]

Then, point z describes a hyperbola having foci at `z_1` and `z_2` and `a in R^(+)`



Important Theorem