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Introduction *-answer-*     Statics is that branch of mechanics which deals with the study of the system of forces in equilibrium.     Matter : Matter is anything which can be perceived by our senses of which can exert, or be acted on, by forces.     Force : Force is anything which changes, or tends to change, the state of rest, or uniform motion, of a body. To specify a force completely four things are necessary they are magnitude, direction, sense and point of application. Force is a vector quantity.     Weight: Weight of a body is the force with which it is attracted towards the centre of the earth. If \[m\] is the mass of the body and \[g,\] the acceleration due to gravity, then its weight \[W=mg\].     Tension and thurst : Whenever a string is used to support a weight or drag a body, there is a force of pull along the string. This force is called tension. Similarly, if some rod be compressed, a force will be exerted. This type of force is called thurst.     Action and reaction : Whenever one body is in contact with another body, there act equal and opposite forces at the point of contact. Such forces are called action and reaction

A function \[\varphi (x)\] is called a primitive or an antiderivative of a function \[f(x)\], if \[\varphi '(x)=f(x).\]     Let \[f\,(x)\] be a function. Then the collection of all its primitives is called the indefinite integral of \[f(x)\] and is denoted by \[\int{f(x)\,dx.}\]     Thus, \[\frac{d}{dx}(\varphi (x)+c)=f(x)\Rightarrow \int{f(x)\,dx}=\varphi (x)+c\]     where \[\varphi (x)\] is primitive of \[f(x)\] and c is an arbitrary constant known as the constant of integration.

(1) The image of a point with respect to the line mirror  The image of \[A({{x}_{1}},\,{{y}_{1}})\] with respect to the line mirror \[ax+by+c=0\] be \[B\,(h,\,\,k)\] is given by,     \[\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}}\]       (2) The image of a point with respect to x-axis : Let \[P(x,y)\] be any point and \[{P}'({x}',\,{y}')\] its image after reflection in the x-axis, then \[{x}'=x\]     \[{y}'=-y,\] (\[\because \] \[{O}'\] is the mid point of P and \[{P}'\])     (3) The image of a point with respect to y-axis : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] its image after reflection in the y-axis,  then \[{x}'=-x\]          \[{y}'=y\],    (\[\because \] \[{O}'\] is the mid point of P and  \[{P}'\])     (4) The image of a point with respect to the origin : Let \[P(x,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection through the origin, then \[{x}'=-x\]                              \[{y}'=-y\],(\[\because \] \[O\] is the mid point of P,\[{P}'\]).     (5) The image of a point with respect to the line \[y=x\] : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection in the line \[y=x\], then \[{x}'=y\]     \[{y}'=x\], (\[\because \]\[{O}'\]is the mid point of P and \[{P}'\]).     (6) The image of a point with respect to the line   \[y=x\tan \theta \] : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection in the line \[y=x\tan \theta \],  then   \[{x}'=x\cos 2\theta +y\sin 2\theta \]    \[{y}'=x\sin 2\theta -y\cos 2\theta \], (\[\because \] \[{O}'\] is the mid point of P and \[{P}'\])      

        Here,   IP = Incident Ray     PN = Normal to the surface     PR = Reflected Ray     Then,  \[\angle IPN=\angle NPR\]     Angle of incidence = Angle of reflection

Three or more lines are said to be concurrent lines if they meet at a point.   First method : Find the point of intersection of any two lines by solving them simultaneously. If the point satisfies the third equation also, then the given lines are concurrent.   Second method : The three lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,\,\,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\] are concurrent if, \[\left| \,\begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}}  \\ \end{matrix}\, \right|\,=\,0\].   Third method : The condition for the lines \[P=0,\,\,Q=0\] and \[R=0\] to be concurrent is that three constants \[a,\,\,b,\,\,c\] (not all zero at the same time) can be obtained such that \[aP+bQ+cR=0\].

Two points \[({{x}_{1}},\,{{y}_{1}})\] and \[({{x}_{2}},\,{{y}_{2}})\] are on the same side or on the opposite side of the straight line \[ax+by+c=0\] according as the values of \[a{{x}_{1}}+b{{y}_{1}}+c\] and \[a{{x}_{2}}+b{{y}_{2}}+c\] are of the same sign or opposite sign.

Let the given line be \[ax+by+c=0\] and observing point is \[({{x}_{1}},\,{{y}_{1}})\], then   (i) If the same sign is found by putting in equation of line \[x={{x}_{1}},\,y={{y}_{1}}\] and \[x=0,\,\,y=0\] then the point \[({{x}_{1}},\,{{y}_{1}})\] is situated on the same side of origin.   (ii) If the opposite sign is found by putting in equation of line \[x={{x}_{1}},\,\,y={{y}_{1}}\] and \[x=0,\,\,y=0\] then the point \[({{x}_{1}},\,{{y}_{1}})\] is situated opposite side to origin.

(1) Distance of a point from a line : The length p of the perpendicular from the point \[({{x}_{1}},\,{{y}_{1}})\] to the line \[ax+by+c=0\] is given by \[p=\frac{|a{{x}_{1}}+b{{y}_{1}}+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\].  

• Length of perpendicular from origin to the line \[ax+by+c=0\] is \[\left| \,\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\, \right|\,\].
• Length of perpendicular from the point \[({{x}_{1}},\,{{y}_{1}})\] to the line \[x\cos \alpha +y\sin \alpha =p\] is \[|{{x}_{1}}\cos \alpha +{{y}_{1}}\sin \alpha -p|\].

  (2) Distance between two parallel lines : Let the two parallel lines be \[ax+by+{{c}_{1}}=0\] and \[ax+by+{{c}_{2}}=0\].   First Method: The distance between the lines is \[d=\frac{|{{c}_{1}}-{{c}_{2}}|}{\sqrt{({{a}^{2}}+{{b}^{2}})}}\].     Second Method: The distance between the lines is \[d=\frac{\lambda }{\sqrt{({{a}^{2}}+{{b}^{2}})}}\],     where (i) \[\lambda =|{{c}_{1}}-{{c}_{2}}|\], if they be on the same side of origin.     (ii) \[\lambda =|{{c}_{1}}|+|{{c}_{2}}|\], if the origin O lies between them.     Third method : Find the coordinates of any point on one of the given line, preferably putting \[x=0\] or \[y=0\]. Then the perpendicular distance of this point from the other line is the required distance between the lines.     Distance between two parallel lines \[ax+by+{{c}_{1}}=0\],\[kax+kby+{{c}_{2}}=0\] is \[\frac{\left| \,{{c}_{1}}-\frac{{{c}_{2}}}{k}\, \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]. Distance between two non parallel lines is always zero.  

The equation of the bisectors of the angles between the lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] are given by,     \[\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\]                                                    .....(i)     Algorithm to find the bisector of the angle containing the origin : Let the equations of the two lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]. To find the bisector of the angle containing the origin, we proceed as follows:     Step I : See whether the constant terms \[{{c}_{1}}\] and \[{{c}_{2}}\] in the equations of two lines positive or not. If not, then multiply both the sides of the equation by \[-1\] to make the constant term positive.     Step II : Now obtain the bisector corresponding to the positive sign i.e.,  \[\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\].     This is the required bisector of the angle containing the origin.     The bisector of the angle containing the origin means the bisector of the angle between the lines which contains the origin within it.     (1) To find the acute and obtuse angle bisectors : Let \[\theta \] be the angle between one of the lines and one of the bisectors given by   (i). Find \[\tan \theta \]. If \[|\tan \theta |<1\], then this bisector is the bisector of acute angle and the other one is the bisector of the obtuse angle.     If \[|\tan \theta |\]> 1, then this bisector is the bisector of obtuse angle and other one is the bisector of the acute angle.     (2) Method to find acute angle bisector and obtuse angle bisector     (i) Make the constant term positive, if not.               (ii) Now determine the sign of the expression \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}\].       (iii) If \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}>0\], then the bisector corresponding to \[''+''\] sign gives the obtuse angle bisector and the bisector corresponding to \[''-''\] sign is the bisector of acute angle between the lines.     (iv) If \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}<0\], then the bisector corresponding to \[''+''\] and \[''-''\] sign given the acute and obtuse angle bisectors respectively.     Bisectors are perpendicular to each other.     If \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}>0\], then the origin lies in obtuse angle and if \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}<0\], then the origin lies in acute angle.  

The equation of the straight lines which pass through a given point \[({{x}_{1}},\,{{y}_{1}})\] and make a given angle \[\alpha \] with given straight line \[y=mx+c\] are, \[y-{{y}_{1}}=\frac{m\,\pm \,\tan \alpha }{1\,\,\mp m\tan \alpha }(x-{{x}_{1}})\].