Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Equation of a line in various forms

(i) Slope-intercept form:

 y=mx+c  is the equation of a straight line whose slope is   m and which makes an intercept  c on the  y  -axis.

Example:

Find the equation of a line with slope  -3  and cutting off an intercept  5   units one negative direction of  y – axis.

 m =-3  and  C =-5

 y=mx+c

 \implies y= -3x -5

 \implies 3x+y+5=0

(ii) Slope-Point form:

Let a line have slope  m and let it pass through a point  Q (x_1,y_1)  then the equation of the line is given by

 y-y_1= m (x-x_1)

Example:

Slope  m=2 and  Q (x_1,y_1) =(3,4) then equation of line is

y-y_1=m (x-x_1)

 \implies y-4=2(x-3)

 \implies y-4=2x-6

 \implies 2x-y-6+4=0

 \implies 2x-y-2=0

(iii) Two-Point form:

A line pass through two given points  P (x_1,y_1) and  Q (x_2,y_2)  . Then the equation of the line is given by

 y-y_1=\left( \dfrac{y_2-y_1}{x_2-x_1} \right)(x-x_1)

where  \dfrac{y_2-y_1}{x_2-x_1} = slope of the line

Example:

 P = (4,9) , \quad Q =(7,2)

Equation of the line is

 y-9=\left( \dfrac{2-9}{7-4} \right)(x-4)

 \implies y-9=-\dfrac{7}{3}(x-4)

 \implies3(y-9)=7(4-x)

 \implies 3y-27=28-7x

 \implies 3y+7x-27-28=0

 \implies 7x+3y-55=0

(iv) Intercept form:

Let  a  line have  x – intercept  a  and  y – intercept   b . Then the equation is

 \dfrac{x}{a}+\dfrac{y}{b}=1

Example:

Equation of the line where a=3, b=4 is

 \dfrac{x}{a}+\dfrac{y}{b}=1

 \implies\dfrac{x}{3}+\dfrac{y}{4}=1

 \implies\dfrac{4x+3y}{12}=1

 \implies 4x+37 -12 =0

(v) Determinant form:

Equation of a line passing through  (x_1, y_1) and  (x_2,y_2) is  \begin{vmatrix}x & y & 1\\ x_1& y_1& 1\\ x_2&y_2 & 1\end{vmatrix}=0

Example:

Let  P (4,5)  and  Q (7, 9) be two points , then equation of the line is   \begin{vmatrix}x & y & 1\\ x_1& y_1& 1\\ x_2&y_2 & 1\end{vmatrix}=0

 \implies\begin{vmatrix}x & y & 1\\ 4 & 5 & 1\\ 7 & 9 & 1\end{vmatrix}=0

 \implies x(5-9)-y (4-7) +1 (36-35)=0

 \implies -4x+3y+1 =0

\implies 4x-3y-1=0

(vi) Perpendicular /Normal Form:

 x \cos \alpha + y \sin \alpha =P ( \text { where} P > 0, O \leq \alpha < 2 \pi) is the equation of the straight line where the length of the perpendicular form the origin   O   on the line is  P and this perpendicular makes an angle  \alpha  with positive  x -axis.  (OP =P)

Example:

Here   P=3, \alpha = 30 ^o

Equation of the line is  x \cos 30^o +y \sin 30^o =3

 \implies x \cdot \dfrac{\sqrt{3}}{2}+y \cdot \dfrac{1}{2}=3

 \implies \sqrt{3} x +y =6

 \implies \sqrt{3} x+ y-6=0

Note:

The normal form of equation  Ax+By+C =0 is  x \cos \alpha + y \sin \alpha= P where

  \cos \alpha= \pm \dfrac{A}{\sqrt{A^2+B^2}}, \quad \sin \alpha=\pm\dfrac{B}{\sqrt{A^2+B^2}}  and 

  P=\pm \dfrac{C}{\sqrt{A^2+B^2}}

(vii) General form:

 ax+by+c=0   is the equation of a straight line in the general form. In this case, slope of line  =-\dfrac{a}{b}  x – intercept  =-\dfrac{c}{a} ,  y – intercept  =-\dfrac{c}{b}

Example:

Find slope ,  x – intercept and y -intercept of the line  2x-3y+5 =0

Hence  a=2, b=-3, c=5

Slope  =-\dfrac{a}{b}=-\dfrac{2}{-3}=\dfrac{2}{3}

 x-intercept  =-\dfrac{c}{a}=-\dfrac{5}{2}

 y-intercept =-\dfrac{c}{b}=-\dfrac{5}{3}

(viii) Parametric Form:

 P(r) -(x, y) =(x_1+r \cos \theta, y_1+r \sin \theta)   or  \dfrac{x-x_1}{\cos \theta}= \dfrac{y-y_1}{\sin \theta}=r is the equation of the line in parametric form, where 'r' is the parameter whose absolute value is the distance  of any point  (x , y) on the line from the fixed point  (x_1, y_1)

\to The above form is derived from slope-point form of the line

 y-y_1=m(x-x_1) where  m =\tan \theta

 \implies y-y_1 = \dfrac{\sin \theta}{\cos \theta} (x-x_1)

Lines Continued:

Consider the equations of lines  L_1 and  L_2   given by

 L_1 : a_1x+b_1y+c_1=0

 L_2 : a_2x+b_2y+c_2=0

Slopes  m_1 and m_2   of L_1 and  L_2 are given by  m_1=-\dfrac{a_1}{b_1}  and  m_2=-\dfrac{a_2}{b_2}   respectively.

Case of Parallel Lines:

The lines L_1 and  L_2 are parallel if  m_1 =m_2  or  \dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}   or    \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{1}{\lambda} 

 \therefore a_2=a_1 \lambda,\quad b_2=b_1 \lambda

Therefore, we can write equation of  L_2   as  \lambda (a_1x+b_1y) +c_2=0

Working Rule:

If a line  L is represented by  ax+by +c=0 , the the equation of a line  L'  , parallel to   L , is given by   ax+by +d=0 .

If , in the equation of  L , either  a  or  b  is zero , then also the equation of   L^1 , parallel to  L , is given by  ax+by+d=0.

Since in this case, both  ax+by +c=0   and  ax+by +d =0 represent lines parallel to the same coordinate axes.

Case of Perpendicular Lines:

The lines represented by  x+a =0   and y+b=0   being respectively vertical and horizontal are mutually perpendicular.

Now consider a line  L : ax+by+c=0 \quad (a \neq 0, \quad b \neq 0)

We can write the equation of a line  L^1 perpendicular to  L , as  L^1 : bx-ay+d=0

The slopes of   L   and   L^1 , respectively  -\dfrac{a}{b}   and  \dfrac{b}{a} . The product of the slopes being  -1  , the lines are  mutually perpendicular .

Working Rule:

To write the equation of  L ^1 , perpendicular to  L  , just interchange the coefficients of  x and  y in the equation of  L and write one of the coefficients by reversin its sign.

Example:

 L : 2x+3y+5 =0

then  L^1 =3x-2y+K=0, K \in R

Hence  L \perp L^1

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