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

Some various forms of the equation of a line

(i) If a line is at a distance  K  and parallel to  x -axis, then the equation of the line is  y= \pm K .

(ii) If a line is parallel to  y -axis at a distance   c from y -axis, the the equation of the lines is  x = \pm c

The position of points is relative to a given line:

Let  L: ax+by+c=0 and two given points  P(x_1 , y_1) \text { and } Q (x_2 ,y_2) .

(i) The two points are on the same side of the straight line  ax+by+c=0   if,  ax_1+by_1+c and  ax_2+by_2+c  have the same sign.

(ii) The two points are on the opposite sides of the line  ax+by+c=0 , if   ax_1+by_1+c   and  ax_2+by_2+c have opposite sign. 

Equation of bisectors of the angle between to intersecting line:



Equation of bisectors   B_1 and B_2 are 


Bisectors of the angle containing a given point (h , k) its interior:

Let the equation of two-line  L_1   and  L_2 be given by

 L_1:a_zx+b_1y+c_1=0, \quad L_2: a_2^1x+b_2^1y+c_2=0

To obtain the equation of the bisector of the angle containing  (h, k ) in its interior, proceed as follows:

(1) First see if  a_1h+b_1K+c_1 and  a^1_2h+b_2^1K+c_2^1   have the same or opposite signs.

(i) If they have the same sign, write a_2=a_2^1, \quad b_2=b_2^1, \quad c_2=c_2^1

(ii) If they have the opposite sign, write  a_2=-a_2^1, \quad b_2=-b_2^1, \quad c_2=-c_2^1

(2) Write the bisector of the angle containing  (h , k ) in its interiro as:



Bisector of the angle, not containing  (h , k) in its interior, is given by  \dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= (-)\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}} ,

Provided   a_1+h+b_1K+c and a_2h+b_2K+c_2 have the same sign. 

Pair of Straight Lines:

Consider the equation  ax^2+2hxy+by^2=0 which is a second degree homogenous equation in  x and  y , can be written as  b\left( \dfrac{y^2}{x^2} \right)+2h \left( \dfrac{y}{x} \right)+a=0 which is  quadratic  in  \dfrac{y}{x} . This gives  \dfrac{y}{x} =\dfrac{-2h\pm \sqrt{4h^2-4ab}}{2b}=\dfrac{-h\pm\sqrt{h^2-ab}}{b}\quad (b \neq0)  or  y =\left( \dfrac{-h+\sqrt{h^2-ab}}{b} \right) x \text { and } y=\left( \dfrac{-h-\sqrt{h^2-ab}}{b} \right) x  which represents a pair of straight lines,  with slopes



Change of Axes (Shifting of Origin)

Let the origin is shifted to a point  O' (h , k) . If  P (x, y) are coordinates of a point referred to old axes and  P' (x , y) are the coordinates of the same points referred to new axes, then  x =x+h , \quad y=y+k


Images of a point with respect to a line:

Let the image of a point  (x_1,  y_1) with respect to  ax+by+c=0  be ( x_2 , y_2) , then  \dfrac{x_2-x_1}{a}=\dfrac{y_2-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}

(i) The image of the point  P (x_1 ,y_1) wrto  x -axis is  Q (x_1 ,y_1)

(ii) The image  P(x_1, y_1) wrto  y-axis is  Q (-x_1,y_1).

(iii) The image of the point P (x_1, y_1) wrto mirror y=x    is  Q (y_1 ,x_1) .

(iv) The image of  P (x_1 , y_1) wrto the line mirror  y =x \tan \theta  is  x =x_1 \cos 2 \theta + y_1 \sin 2 \theta

 y= 2\sin 2 \theta-y_1 \cos 2 \theta

(v) The image of  P(x_1,y_1) wrto origin is (-x_1,y_1)

(vi) The length of perpendicular from a point  (x_1,y_1) to a line  ax+by +c =0 is  \left| \dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \right|

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