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G Gautam harsolia
Given equation id line is We can rewrite it as Now, the distance of the line   from the point    is given by  Similarly, The distance of the line   from the point    is given by                                                                                                  Hence  proved

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G Gautam harsolia
point of intersection of lines    and   (junction) is   Now, person reaches to path   in least time  when it follow the path perpendicular to it  Now, Slope of line  is ,  let the slope of line perpendicular to it is , m Then, Now, equation of line passing through point   and with slope   is  Therefore, the required equation of line is

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G Gautam harsolia
From the figure above we can say that The slope of line AC  Therefore, Similarly, The slope of line AB  Therefore, Now, from equation (i)  and (ii) we will get Therefore, the coordinates of .  is

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G Gautam harsolia
Let's take the point  which is equidistance from  parallel lines     and   Therefore,                                                                               It is that    Therefore, Now, case (i) Therefore, this case is not possible  Case (ii) Therefore, the required equation of the line is

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G Gautam harsolia
Given the equation of line are Now, perpendicular distances of a variable point    from the lines are                                                                                        Now, it is given that Therefore,                                                                    Which is the equation of the line  Hence proved

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G Gautam harsolia
Given equation of lines are  Now, it is given that line (i) and (ii)  are equally inclined to the line (iii) Slope of line   is  ,   Slope of line  is ,  Slope of line  is ,  Now, we know that Now,               and                   It is given that  Therefore, Now, if      Then, Which is not  possible Now,  if  Then, Therefore, the value of  m is

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G Gautam harsolia
Let point  is the image of point  w.r.t. to line  line   is perpendicular bisector of line joining points    and  Slope of line  ,  Slope of   line joining points    and   is  ,  Now, Point of intersection is the midpoint of line  joining points    and  Therefore, Point of intersection is   Point  also  satisfy the line   Therefore, On solving equation (i) and (ii) we will...

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G Gautam harsolia
Slope of line OA and OB are negative times inverse of each other  Slope of line OA is  ,   Slope of line OB is ,  Now, Now, for a given value of  m we get these equations If

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G Gautam harsolia
Let  be the point of intersection it lies on line   Therefore,  Distance of point  from  is 3 Therefore, Square both the sides and put value from equation (i) When        point is   and When       point is  Now, slope of line joining point    and    is  Therefore, line is parallel to x-axis                       -(i) or slope of line joining point   and   Therefore, line is...

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G Gautam harsolia
point  lies on line  Now, point of intersection of lines      and      is  Now, we know that the distance between two point is given by  Therefore, the distance of the line    from the point    along the line   is

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G Gautam harsolia
Equation of line joining    and   is Now, point of intersection of lines   and     is    Now, let's suppose point divides the line  segment   joining    and     in   Then, Therefore, the line joining    and    is divided by the line    in ratio

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G Gautam harsolia
Slope of line  is m Let  the slope of other line is m' It is given that both the line makes  an angle   with each other Therefore, Now, equation of line passing through origin (0,0) and with slope    is Hence proved

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G Gautam harsolia
Point of intersection of the lines    and   is   We know that the intercept form of the line is  It is given that line make equal intercepts on x and  y axis Therefore, a = b Now, the equation reduces to          -(i) It passes through point     Therefore, Put the value of a in equation (i) we will get Therefore, equation of line is

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G Gautam harsolia
Given the equation of the line is The slope of line  ,  Let the slope of the other line is,  Now, it is given that both the lines make an angle  with each other  Therefore, Now, Case (i)                        Equation of line passing through the point    and  with slope  Case (ii) Equation of line passing through the point    and  with slope 3  is Therefore,...

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G Gautam harsolia
Concurrent lines means they all intersect at the same point Now, given equation of lines are  Point of intersection  of equation (i) and (ii)   Now, lines are concurrent which means point   also satisfy equation (iii) Therefore, Hence proved

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G Gautam harsolia
Point of intersection of lines  and  is   Now,  must satisfy equation  Therefore, Therefore, the value of p is

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G Gautam harsolia
Given equations of lines are The point if intersection of (i)  and (ii) is  (0,0) The point if intersection of (ii)  and (iii) is  (k,-k) The point if intersection of (i)  and (iii) is  (k,k) Therefore, the vertices of triangle formed by three lines are  Now, we know that area of triangle whose vertices are   is Therefore, area of triangle  is

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G Gautam harsolia
given equation of line is we can rewrite it as Slope of line  ,  Let the Slope of perpendicular line is m Now, the ponit of intersection of   and   is    Equation of line passing through point  and with slope   is Therefore, equation of line is

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G Gautam harsolia
Point of intersection of the lines    and   It is given that this line is parallel to y - axis i.e.  which means their slopes are equal Slope of  is , Let the Slope of line passing through point  is m Then, Now, equation of line passing through point  and with slope  is Therefore, equation of line is

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G Gautam harsolia
Equation of line passing through the points     and   is                                                                                          Now, distance from origin(0,0) is
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