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Straight Line - Best Maths Practice Material

Straight Line

Coordinate Geometry

Coordinate geometry is also known as analytical geometry or coordinate geometry. Co-ordinate geometry is the study of geometry by algebraic methods, like studying various geometrical curves and properties with the help of algebraic equations.

Coordinate System:: Co-ordinate system is also called a rectangular coordinate system i.e Plane is divided into four quadrants with the help of two mutually perpendicular reference lines, called as X-axis and Y-axis.

There are two ways in which we can uniquely fix a point in the coordinate system, also called a 2-Dimension XY Co-ordinate system.

Different forms of Co-ordinate system are:-

cordinate geometry
Co-ordinate System

(1) Cartesian form:- Let us consider a point $P(x_1, y_1)$ in a rectangular coordinate system, where $x_1$ is directed distance of point $P$ from Y-axis, if $x_1 \gt 0$ then the distance is measured towards the right side of Y-axis (i.e toward the positive direction of X-axis ) and if $x_1 \lt 0$ then the distance is measured towards the left side of Y-axis ( i.e toward the negative direction of X-axis ).

Similarly, if $y_1$ is directed distance measured from X-axis if $y_1 \gt 0$ then the distance is measured above the X-axis ( i.e toward the positive direction of Y-axis ), if $y_1 \lt 0$ then the distance is measured below the X-axis ( i.e towards the negative direction of Y-axis ).

(2) Polar Coordinate System:: Another method to represent a point uniquely on the coordinate system is a polar coordinate system, where point P is represented as the distance from reference / fixed point origin and direction represented by the angle with respect to the positive direction of the x-axis. The point P is represented as $P(r,\theta )$ in a polar co-ordinate system.

Polar Coordinate System
Polar form of Co-ordinate System $r\ge 0 $ and $\theta \in (-\pi , \pi ]$

The relation between Polar and Rectangular Cartesian System.

Let $P(h,k)$ be a point at a distance of $r$ units from origin O and OP makes an angle of $\theta $ with positive direction of X-axis then

$$r=\sqrt{h^2+k^2}\,\,,\,\,h=r Cos\theta \;,\; k=r Sin\theta $$

Illustration 1:- Transform the Cartesian into polar and polar into Cartesian form

(i) $(- \sqrt{3},1)$(ii) $ (3,4) $(iii) $ (4,\frac{7\pi }{6}) $(iv) $ (\sqrt{2} , \frac{-\pi }{3}) $
Solution :-(i)$x=-\sqrt{3}=r.Cos\theta $and $y=1=r.Sin\theta $
 Where $r=\sqrt{x^2+y^2}=\sqrt{3+1}=2$$tan\theta = -\frac{1}{\sqrt{3}}$
 $\therefore \theta=\frac{5\pi }{6}$ 
Hence polar form of $(-\sqrt{3},1)$ is$(2,\frac{5\pi }{6})$
(ii)$x=3=rCos\theta $and $y=4=rSin\theta$
 Where $r=\sqrt{x^2+y^2}=\sqrt{9+16}=5$$tan\theta = \frac{4}{3}$
 $r =\sqrt{9+16}=5$$\therefore \theta=tan^{-1}(\frac{4}{3})$
 Hence polar form of $(3,4) \,\,\,is\,\,\,(5,tan^{-1}(\frac{4}{3}))$
(iii)$r=4$$\theta = \frac{7\pi }{6}$
 $x=rCos\theta $$y=rSin\theta $
 $x=4.Cos(\frac{7\pi }{6})=-4.\frac{\sqrt{3}}{2}$$y=4.Sin(\frac{7\pi }{6})=-4.(\frac{1}{2})$
 $x=-2\sqrt{3}$$y=-2$
 Hence Cartesian form is $(-2\sqrt{3},-2)$
(iv)$r=\sqrt{2}$$\theta = -\frac{\pi }{3}$
 $x=rCos\theta $$y=rSin\theta $
 $x=\sqrt{2}.Cos(-\frac{\pi }{3})=\sqrt{2}.(\frac{1}{2})$$y=\sqrt{2}.Sin(-\frac{\pi }{3})=-\sqrt{2}.\frac{\sqrt{3}}{2}$
 $x=\frac{1}{\sqrt{2}}$$y=-\sqrt{\frac{3}{2}}$
 Hence Cartesian form is$(\frac{1}{\sqrt{2}}, -\sqrt{\frac{3}{2}} )$

Illustration 2 :- Find distance between the points $(a\,Cos\alpha , a\,Sin\alpha )$ and $(a\,Cos\beta , a\,Sin\beta ) $ where $(a \gt 0) $.

Solution :- Let point A and B is $(a\,Cos\alpha , a\,Sin\alpha ) $ and $ (a\,Cos\beta , a\, Sin\beta )$ respectively

$$AB=\sqrt{(a\,Cos\alpha-a\,Cos\beta)^2+(a\,Sin\alpha – a\,Sin\beta )^2}$$

$$\therefore AB=a\sqrt{2(1-Cos(\alpha – \beta))}$$

$$\therefore AB=a\sqrt{4Sin^2(\frac{\alpha -\beta}{2})}$$

$$AB=2a \vert Sin(\frac{\alpha – \beta}{2}) \vert $$

Distance Formulae.

Distance between point $A(x_1,y_1)$ and $B(x_2,y_2)$ in Cartesian form

$\Delta ABM$ is a right angle triangle

$\Delta ABM$ is a right angle triangle with $AM=\vert x_2-x_1 \vert $ and $BM=\vert y_2 – y_1 \vert $

$$\therefore AB=\sqrt{AM^2+BM^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

NOTE ::- (1) If AB is parallel to $X-axis$ then $AB=\sqrt{(x_2-x_1)^2}=\vert x_2 – x_1 \vert $

(2) If AB is parallel to $Y-axis$ then $AB=\sqrt{(y_2-y_1)^2}=\vert y_2 – y_1 \vert $

Distance between points in polar form.

Distance between points $A(r_1,\theta_1)$ and $B(r_2,\theta_2)$

Let point A and B are $(r_1 , \theta_1)$ and $(r_2 , \theta_2)$ respectively.

In $\Delta OAB $ use Cosine Rule we get ,

$$Cos(\theta_1 – \theta_2)=(\frac{r_1^2+r_2^2-(AB)^2}{2r_1r_2})$$

$$\therefore AB=\sqrt{r_1^2+r_2^2-2r_1r_2Cos(\theta_1 – \theta_2) } $$

Section formulae.

Let point P divide the line segment $AB$ in the ratio $m:n$ ( i.e $\frac{AP}{BP}=\frac{m}{n} $).

If m:n$\neq$1:1 always there exist two sections of line segment $AB$ one where P lies between A, B and another case is when point P lie outside AB.

CASE I :- INTERNAL SECTION

When point P divide A B internally in the ratio m:n is called case of internal section

Internal Section :: Point P divide line joining A B internally in ratio m : n

$$\Delta AMP \sim \Delta PNB \,\,\,\, ( By \, AAA\, Criterian\,)$$

$$\therefore \frac{AP}{BP}=\frac{AM}{PN}=\frac{PM}{BN}=\frac{m}{n}$$

$$\Rightarrow \frac{x-x_1}{x_2-x}=\frac{m}{n}$$

$$\Rightarrow nx-nx_1=mx_2-mx$$

$$\therefore (m+n)x=mx_2+nx_1$$

$$\Rightarrow x=\cfrac{mx_2+nx_1}{m+n}$$

Similarly for $y$ co-ordinate $y=\cfrac{my_2+ny_1}{m+n}$

CASE II :- EXTERNAL SECTION

When point P divide A B externally in the ratio $m:n$ is case of external section

External Section :: Point P divide line joining A B externally in ratio m : n

Let co-ordinates of point A , B are $(x_1,y_1)\,\,and\,\,(x_2,y_2)$ and point $P(x,y)$ be a point that divide line segment A B externally such that $\cfrac{AP}{BP}=\cfrac{m}{n}$

$$\Delta AMP \sim \Delta BNP \;\;(By\,AAA\,Criterian\,)$$

$$\therefore \frac{AP}{BP}=\frac{AM}{BN}=\frac{PM}{PN}=\frac{m}{n}$$

$$\Rightarrow \frac{x-x_1}{x-x_2}=\frac{m}{n}$$

$$\Rightarrow nx-nx_1=mx-mx_2$$ $$\therefore (m-n)x=mx_2-nx_1$$ $$\Rightarrow x=\frac{mx_2-nx_1}{m-n}$$

Similarly for $y$ co-ordinate $y=\frac{my_2-ny_1}{m-n}$

 

CASE III :- HARMONIC CONJUGATE

Harmonic Conjugate where $PS:RS=\lambda :1 \,,\,PQ:QR=\lambda :1$

The point $Q$ and $S$ divide $PR$ internally and externally respectively in the same ratio , then point $Q$ and $S$ are harmonic conjugate of each other with respect to $P$ and $R$. $$i.e\;\;\frac{PS}{RS}=\frac{\lambda}{1}\;,\;\frac{PQ}{QR}=\frac{\lambda}{1}$$

If $Q$ and $S$ are harmonic conjugate of $P$ and $R$ then $P$ and $R$ are harmonic conjugate of $Q$ and $S$.

If $Q$ and $S$ are harmonic conjugates of $P$ and $R$ then $PQ\,,\, PR\,,\, PS$ is in Harmonic progression.

NOTE : – (1) Mid-point of point $A(x_1,y_1)$ and $B(x_2,y_2)$ is $\Bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Bigr)$

(2) Centroid (G) of $\Delta ABC$ where co-ordinates of point $A\,,\,B\,,\,C$ are $(x_1,y_1)\,,\,(x_2,y_2)\,,\,(x_3,y_3)$ respectively is $\Bigl(\frac{x_1+x_2+x_3}{3}\,,\,\frac{y_1+y_2+y_3}{3}\Bigr)$

(3) Centroid (G) divide orthocentre (H) and circumcentre (O) in the ratio $HG:GO=2:1$.

(4) In $\Delta ABC\;,\;AD\,,\,BE\,,\,CF$ are internal angle bisector of angle $A\,,\,B\,,\,C$ and $I_1I_2\,,\,I_2I_3\,,\,I_3I_1$ are external angle bisector of angles $C\,,\,A\,,\,B$

In-centre and Ex=centre of triangle ABC

In $\Delta ABC$ if sides $AB=c\,,\,BC=a\,,\,AC=b$ and co-ordinates of points $A(x_1,y_1)\,,\,B(x_2,y_2)\,,\,C(x_3,y_3)$

In-centre $I$ is points of intersection of internal angle bisectors and Ex-centre $I_1\,,\,I_2\,,\,I_3$ are points of intersection of two external angle bisector with third internal angle bisector.$$\frac{AI}{ID}=\frac{b+c}{a}\;,\;\frac{BI}{IE}=\frac{a+c}{b}\;,\;\frac{CI}{IF}=\frac{a+b}{c}$$ $$\frac{AI_1}{I_1D}=\frac{b+c}{a}\;,\;\frac{BI_2}{I_2E}=\frac{a+c}{b}\;,\;\frac{CI_3}{I_3F}=\frac{a+b}{c}$$

The co-ordinate In-centre are $$I=\Bigl(\frac{a.x_1+b.x_2+c.x_3}{a+b+c}, \frac{a.y_1+b.y_2+c.y_3}{a+b+c}\Bigr)$$ The co-ordinate Ex-centre are $$I_1=\Bigl(\frac{-a.x_1+b.x_2+c.x_3}{-a+b+c}, \frac{-a.y_1+b.y_2+c.y_3}{-a+b+c}\Bigr)$$ $$I_2=\Bigl(\frac{a.x_1-b.x_2+c.x_3}{a-b+c}, \frac{a.y_1-b.y_2+c.y_3}{a-b+c}\Bigr)$$ $$I_3=\Bigl(\frac{a.x_1+b.x_2-c.x_3}{a+b-c}, \frac{a.y_1+b.y_2-c.y_3}{a+b-c}\Bigr)$$

Illustration 3:- The four points $P(\alpha , 0)\,,\,Q(\beta ,0)\,,\,R(\gamma , 0)\,,\,S(\delta , 0)$ are such that $\alpha \,,\,\beta $ are the roots of equation $ax^2+2hx+b=0$ and $\gamma \,,\,\delta $ are those of equation $a’x^2+2h’x+b’=0$ . Show that the sum of the ratios in which $R$ and $S$ divide $P Q$ is zero , if $ab’+a’b=2hh’$.

Solution :- Given that $\alpha \,,\, \beta$ is the root of equation $ax^2+2hx+b=0$ $$\therefore \alpha+\beta =-\frac{2h}{a}\,\,and\,\,\alpha\beta=\frac{b}{a}$$

Given that $\gamma \,,\, \delta$ is the root of equation $a’x^2+2h’x+b’=0$ $$\therefore \gamma+\delta =-\frac{2h’}{a’}\,\,and\,\,\gamma\delta=\frac{b’}{a’}$$

Let point $R$ divide $P Q$ in the ratio $m:1$ then $$\gamma = \frac{m\beta +\alpha}{m+1} \,\,\Rightarrow m=\frac{\gamma – \alpha}{\beta – \gamma} $$

and Let point $S$ divide $P Q$ in the ratio $n:1$ then $$\delta = \frac{n\beta +\alpha}{n+1} \,\,\Rightarrow n=\frac{\delta – \alpha}{\beta – \delta } $$

Given that $ m \,+\, n=0$

$$\Rightarrow \frac{\gamma – \alpha}{\beta – \gamma} + \frac{\delta – \alpha}{\beta – \delta } = 0$$ $$\Rightarrow (\alpha + \beta )(\gamma + \delta ) -2\alpha\beta – 2\gamma\delta = 0 $$

$$ ab’+a’b=2hh’$$

Illustration 4:- The vertices of triangle are $A\,(a\,,\,a.tan\alpha )\,,\,B\,(b\,,b.tan\beta )\,and\, C(c\,,\,c.tan\gamma)$. If circumcentre of $\Delta ABC$ co-insides with origin and $(h,k)$ is orthocentre . Then find $k:h$.

Solution :- Since origin is circumcentre of $\Delta ABC$ then

$$\Rightarrow OA=OB=OC=R $$ $$\therefore a^2(1+tan^2\alpha)= b^2(1+tan^2\beta)= c^2(1+tan^2\gamma)=R^2$$ $$\therefore a\,Sec\alpha= b\,Sec\beta= c\,Sec\gamma= R$$ $$\therefore a=R\,Cos\alpha \,,\, b=R\,Cos\beta \,,\, c=R\,Cos\gamma $$

Hence vertices of $\Delta ABC$ are $A(RCos\alpha , RSin\alpha )\,,\, B(RCos\beta , RSin\beta )\,,\, C(RCos\gamma , RSin\gamma )$

$$\therefore Centroid \,\,G=\Bigl(\frac{R}{3}\sum Cos\alpha\,,\,\frac{R}{3}\sum Sin\alpha \Bigr)$$

We know that centroid divide orthocentre and circumcentre in ratio $2:1$ internally .

Hence $H=(R\sum Cos\alpha , R\sum Sin\alpha )=H(h,k)$

$$\Rightarrow \frac{k}{h}=\frac{Sin\alpha\,+\,Sin\beta\,+\,Sin\gamma}{Cos\alpha\,+\,Cos\beta\,+\,Cos\gamma}$$

Area of Triangle

Let $A(x_1,y_1)\,,\,B(x_2,y_2)\,,\,C(x_3,y_3)$ are the vertices of $\Delta ABC$ then $$ $$ Area of $\Delta ABC\,=\,\vert $ Area of trapezium ABQP + Area of trapezium BCRQ – Area of trapezium ACRP $\vert $.

Area of $\Delta ABC \,=\,\frac{1}{2} \vert (y_1+y_2)(x_2-x_1)+(y_2+y_3)(x_3-x_2)-(y_3+y_1)(x_3-x_1)\vert$

$$=\frac{1}{2}\vert x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\vert$$

$$=\frac{1}{2} \begin{Vmatrix} 1&x_1&y_1\\ 1& x_2&y_2\\ 1& x_3 & y_3\\ \end{Vmatrix}$$

NOTE :- (1) Area of n – sided polygon whose co-ordinate of vertices are $$A_i=(x_i,y_i)\,\,,\,\,\forall\, i=1,2,3,4,..,n$$ is given by $ \frac{1}{2}\Bigl( \begin{vmatrix} x_1&y_1\\ x_2&y_2\\ \end{vmatrix} + \begin{vmatrix} x_2&y_2\\ x_3&y_3\\ \end{vmatrix} + \begin{vmatrix} x_3&y_3\\ x_4&y_4\ \end{vmatrix} +….+ \begin{vmatrix} x_n&y_n\\ x_1&y_1\\ \end{vmatrix} \Bigr)$

(2) Condition for three points $(x_1,y_1)\,,\,(x_2,y_2)\,,\,(x_3,y_3)$ to be col-linear is Area of triangle is Zero. $$\Rightarrow \begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\\ \end{vmatrix}=0$$

Illustration 5:- The vertices of $\Delta ABC $ are $A\,(a\,,\,b)\,,\,B\,(1\,,\,2 )\;and\; C(2\,,\,3)\;,where\;a\,,\,b \in I$. Point $A$ lie on line $y\,=\,2x\,+\,3$ . If area of $\Delta ABC$ be such that $\lfloor \Delta \rfloor\,=\,2$ , where $\lfloor . \rfloor $ denotes the greatest integral function ( floor function ) . Find all possible co-ordinates of point $A$.

Illustration 6:- If $P\,(3\,,\,1)\,,\,Q\,(6\,,\,5 )\;and\; R(a\,,\,b)$ are three points such that $\angle PRQ = 90^0$ and area of $\Delta PRQ=7$ . Then find number of possible co-ordinates of $R$.

Illustration 7:- If $m_1$ and $m_2$ are the roots of the equation $x^2+(\sqrt{3}+2)x+(\sqrt{3}-1)=0$ . Find area of the triangle formed by lines $y=m_1 x\;,\;y=m_2 x$ and $y=c$ .

Illustration 8:- If $(x_1,y_1)\,,\,(x_2,y_2)\,and\,(x_3,y_3)$ are vertices of triangle , such that $x_1\,,\,x_2\,,\,x_3\,and\,y_1\,,\,y_2\,,\,y_3$ are two G.P’s having common ratio $r_1\,are\,r_2$ . Find area of triangle.

Illustration 9:- If point $A$ divides the join of $P(-5,1)$ and $Q(3,5)$ in the ratio $\lambda :1$ . Find the quadratic equation whose roots are values of $\lambda$ such that area of $\Delta ABC$ where $B(1,5)$ and $C(7,-2)$ is $2$ square units.

Locus

Locus of a point is the path traced by a moving/variable point under the given conditions.

Steps to find Locus.

(1) Assume the variable/ moving point be $P(h,k)$ whose locus to be determined.

(2) Express given condition in mathematical express, involving $h$ and $k$.

(3) Eliminate arbitrary parameter / if any used in the formation of expression in step 2.

(4) The expression after step 3 is independent of arbitrary unknown/parameter and now involve only in terms of $h,k$ and fixed/known constants. Now replace $(h,k)$ with $(x,y)$.

Straight Line

Locus of a point such that area of triangle formed by variable / moving (or General point ) point $P(x,y)$ and two fixed points $A(x_1,y_1)\,,\,B(x_2,y_2)$ is zero. i.e $\begin{vmatrix} x&y&1\\x_1&y_1&1\\x_2&y_2&1\\ \end{vmatrix}=0$

INCLINATION OF LINE.

Inclination of a line is the angle $\theta $ which line makes with positive direction of x-axis measured in anticlockwise direction. $i.e\;\theta \in [0,\pi)$

Inclination of line $theta \in [0,\pi)$

SLOPE OF A LINE

Slope is tangent of angle which the line makes with positive direction of x-axis .

i.e Slope of line $m=tan \theta $

(1) Slope of line parallel to x-axis is $m=0$.

(2) Slope of line parallel to y-axis is not-defined.

The slope of Line / Line segment joining two fixed points.

Slope of line joining point A and B is $m=\cfrac{y_2-y_1}{x_2-x_1}$

In the $\Delta ABM$ the co-ordinates of points $A(x_1,y_1)\,,\,B(x_2,y_2)$ and $M(x_2,y_1)$ .

Length of sides $AM=(x_2 – x_1)\;,\;BM=(y_2-y_1)$ Hence Slope of line AB is $$m_{AB}=tan \theta =\frac{y_2-y_1}{x_2-x_1}$$

Equation of straight line in different forms.

(1) General form of line

Equation of line in general form is $ax+by+c=0$

Let point $P(x,y)$ divide the line joining two points $A(x_1,y_1)$ and $B(x_2,y_2)$ on the line $ax+by+c=0 $ in $k:1$ where $k \in R-\{-1\}$. If for all $k$ point $P=(x,y)=(\cfrac{kx_2+x_1}{k+1},\cfrac{ky_2+y_1}{k+1})$ satisfy the equation then the locus / curve represent a line.

Given point $A(x_1,y_2) \,,\,B(x_2,y_2)$ lie on line $ax+by+c=0$ , Hence

$$ ax_1+by_1+c=0\,—–(1)$$ $$ax_2+by_2+c=0\,—–(2)$$

Put co-ordinate of point $P$ in equation of line we get

$$a(\frac{kx_2+x_1}{k+1})+ b(\frac{ky_2+y_1}{k+1})+c+0$$

$$\Rightarrow (ax_1+by_1+c)+k(ax_2+by_2+c)=0$$

Using (1) and (2) we get $k \in R-\{-1\}$

Hence equation $ax+by+c=0$ represent the equation of line.

(2) Equation of line in slope and point form

Equation of line $L$ with slope $m$ and passing through point $A(x_1,y_1)$ .

Slope of line $A(x_1,y_1)$ and $P(x,y)$ is $m_{AP}=\frac{y-y_1}{x-x_1}$

Hence line in slope and point form is $ y-y_1=m (x-x_1)$

(3) Equation of line in two point form .

Locus of point $P(x,y)$ such that $P\,,\,A\,,\,B$ are col-linear , where co-ordinate of point $A\,,\,B$ are $(x_1,y_1)\,,\,(x_2,y_2)$ respectively.

$$i.e\;m_{AP}=m_{AB}$$

$$\Rightarrow \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$$

$$\therefore y-y_1=\Bigl(\frac{y_2-y_1}{x_2-x_1}\Bigr)(x-x_1)$$

(4) Equation of line in intercept form

Equation of line with x-intercept and y-intercept as $a\,,\,b$ respectively is $$\frac{x}{a}+\frac{y}{b}=1\;,where\;a,b\neq 0$$

(5) Equation of line in slope and y-intercept form.

Equation of line with slope of line $’m’$ and y-intercept of $c$ on y-axis is $y=mx+c$.

(6) A normal form of a line

Let $p$ be the length of the perpendicular from origin $O$ to the line $L$ and $\alpha \in [0,2\pi)$ be the inclination of perpendicular ( Normal ).

In $\Delta OMA\;,\;Cos\alpha = \frac{OM}{OA}\;\Rightarrow OA = pSec\alpha $

In $\Delta OMB\;,\;Sin\alpha = \frac{OM}{OB}\;\Rightarrow OB = pCosec\alpha $

where $p\gt 0$ is perpendicular from origin on line $L$

$\therefore $ Equation of line $L$ is $\cfrac{x}{pSec\alpha}+\cfrac{y}{pCosec\alpha}=1 \;\;\Rightarrow (Cos\alpha)x+(Sin\alpha)y=p$

(7) Parametric form of line (Symmetric form)

Let line passes through the point $A(x_1,y_1)$ and $P(x,y)$ is a variable point on the line such that distance between $AP=r$ unit.

where $r$ is directed distance if and $r\in R$ , if one of direction of measurement with respect to point $A$ is taken to be positive other direction will be negative.

$\therefore \,in\;\Delta APM \;,\; Sin\theta=\cfrac{y-y_1}{r}\;and\; Cos\theta =\cfrac{x-x_1}{r}\;\;\Rightarrow \cfrac{x-x_1}{Cos\theta}=\cfrac{y-y_1}{Sin\theta}=r$

Hence any point $P(x,y)$ on line is given by $x=rCos\theta +x_1\,,\,y=rSin\theta+y_1 $

Reduction of line from General Form of line to different forms.

(1) General form to slope and intercept form.

The equation of line $ax+by+c=0$ can be written as $$y=-\frac{a}{b}x+(-\frac{c}{b})\;where\;m=-\frac{a}{b}\;,\;y-intercept =-\frac{c}{b}$$

NOTE :- Slope of line $ax+by+c=0$ is given by $$m=-\frac{a}{b}=-(\frac{Coefficient\,of\,x}{Coefficient\,of\,y})$$

(2) General form to Intercept form.

The equation of line $ax+by+c=0$ can be written as $(\frac{x}{(-\cfrac{c}{a})})+(\frac{y}{(-\cfrac{c}{b})})=1$

Hence x-intercept and y-intercept of line $ax+by+c=0$ is $-\cfrac{c}{a}$ and $-\cfrac{c}{b}$ respectively.

(3) General form to Normal form of line.

$ax+by+c=0$ can be written as $ax+by=-c\;,if\;c\lt 0$

$$\therefore \cfrac{a}{\sqrt{a^2+b^2}}x+\cfrac{b}{\sqrt{a^2+b^2}}y=\cfrac{-c}{\sqrt{a^2+b^2}}$$

$$Where\;Cos\alpha=\cfrac{a}{\sqrt{a^2+b^2}}\;,\;Sin\alpha =\cfrac{b}{\sqrt{a^2+b^2}}\;and\;p=\cfrac{-c}{\sqrt{a^2+b^2}}$$

Angle between the lines.

Angle $\angle ABC =\theta_2 \;,\angle BAC =\theta \;and \;\angle ACX=\theta_1$

 

Let lines $L_1\,(AC)$ and $L_2 \,(AB)$ makes an angle of $\theta_1\;,\;\theta_2$ with positive direction of x-axis respectively.

Using Exterior Angle of a triangle is equal to the sum of opposite angles.

We get $\theta = \theta_1-\theta_2\,—–(1) $

where as $\theta$ is angle between lines $L_1$ and $L_2$. Corresponding to $\theta$ there exist another angle between the lines i.e $\pi – \theta$.

From (1) $\tan\theta =tan(\theta_1-\theta_2)=\cfrac{tan\theta_1-tan\theta_2}{1+tan\theta_1 tan\theta_2}$

Since $\theta$ and $\pi- \theta$ are angles between lines $L_1$ and $L_2$ , and $\vert tan\theta \vert =\vert tan(\pi – \theta) \vert $

Hence acute angle between lines is given by $ tan\theta =\vert \cfrac{tan\theta_1-tan\theta_2}{1+tan\theta_1 tan\theta_2} \vert $

NOTE :- (1) Condition for lines to be parallel i.e $\theta\,=\,0$

$$\Rightarrow m_1=m_2$$

Let equation of lines are $\begin{matrix} L_1&:&a_1x+b_1y+c_1=0 \\ L_2&:&a_2x+b_2y+c_2=0 \\ \end{matrix}$

$$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}$$

(2) Condition for lines to be perpendicular

$$\Rightarrow \theta =\frac{\pi}{2}$$

$$\Rightarrow m_1.m_2=-1 $$ i. product of slopes of two lines is $-1$

Hence $ a_1.a_2 + b_1.b_2=0$

Position of a point with respect to a line.

Every line in a plane divides the plane into two regions ( also called as sides ). Any point on the plane which does not lie on a line, lie in one of the regions with respect to the line.

(1) Condition for points $(x_1,y_1)$ and $(x_2,y_2)$ to lie on same side of a line.

Let point $A(x_1,y_2)$ and $B(x_2,y_2)$ lie on same side of a line $L:ax+by+c=0$

If line joining $AB$ is parallel to line $L$ then point $A$ and $B$ lie on same side with respect to line.

Points $A$ and $B$ lie on same side of line L. Point P divide line joining AB Externally i.e $K<0$

If line joining $AB$ is not parallel to line L , then line joining $AB$ intersects the line $L$ at point $P$ , where point $P$ divide line joining $AB$ externally in the ratio of $AP:BP=K:1$

Co-ordinates of point p is given by $P=(\cfrac{Kx_2+x_1}{K+1}\,,\,\cfrac{Ky_2+y_1}{K+1})$

Point P lie on line L . $\therefore a(\cfrac{Kx_2+x_1}{K+1})+b( \cfrac{Ky_2+y_1}{K+1})+c=0$

$$\Rightarrow k=-\Bigl(\cfrac{ax_1+by_1+c}{ax_2+by_2+c}\Bigr)\;,where\,\,k\lt0\,\,( Case\, of \,external\, section ).$$

$$\therefore \Bigl(\cfrac{ax_1+by_1+c}{ax_2+by_2+c}\Bigr)\gt 0$$

Hence $(ax_1+by_1+c)(ax_2+by_2+c)\gt 0$ is the condition for two points $A(x_1,y_1)$ and $B(x_2,y_2)$ lie on same side with respect to the line L.

(2) Condition for points $(x_1,y_1)$ and $(x_2,y_2)$ to lie on opposite side of a line.

Points on opposite sides of given line
If points $A,B$ lie on opposite side of the line $L$ then point P divide line joining $AB$ internally in the ratio of $AP:PB=K:1$

Let point $A(x_1,y_2)$ and $B(x_2,y_2)$ lie on opposite side of a line $L:ax+by+c=0$ . If line joining $AB$ intersects the line $L$ at point $P$ , such that point $P$ divide the line joining Internally in the ratio of $AP:BP=K:1$

Co-ordinates of point p is given by $P=(\cfrac{Kx_2+x_1}{K+1}\,,\,\cfrac{Ky_2+y_1}{K+1})$

Point P lie on line L . $\therefore a(\cfrac{Kx_2+x_1}{K+1})+b( \cfrac{Ky_2+y_1}{K+1})+c=0$

$$\Rightarrow k=-\Bigl(\cfrac{ax_1+by_1+c}{ax_2+by_2+c}\Bigr)\;,where\,\,k\gt 0\,\,( Case\, of \,internal \,section ).$$

$$\therefore \Bigl(\cfrac{ax_1+by_1+c}{ax_2+by_2+c}\Bigr)\lt 0$$

Hence $(ax_1+by_1+c)(ax_2+by_2+c)\lt 0$ is the condition for two points $A(x_1,y_1)$ and $B(x_2,y_2)$ lie on opposite side with respect to the line L.