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Q: 23     Prove that the product of the lengths of the perpendiculars drawn from the points  and    to the line     is .

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

Q : 24    A person standing at the junction (crossing) of two straight paths represented by the equations    and    wants to reach the path whose equation is    in the least time. Find equation of the path that  he should follow.

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

Q : 22     A ray of light passing through the point    reflects on the -axis at point and the reflected ray passes through the point . Find the coordinates of

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

Q : 21     Find equation of the line which is equidistant from parallel lines     and  .

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

Q: 20    If the sum of the perpendicular distances of a variable point    from the lines    and    is always  .  Show that    must move on a line.

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

Q : 19     If the lines    and    are equally inclined to the line  , find the value of

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

Q : 18    Find the image of the point    with respect to the line    assuming the  line to be a plane mirror.

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...

Q : 17     The hypotenuse of a right angled triangle has its ends at the points    and    Find an equation of the legs (perpendicular sides) of the triangle.

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

Q: 16     Find the direction in which a straight line must be drawn through the point    so that its point of intersection with the line    may be at a distance of units from this point.

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...

Q : 15     Find the distance of the line    from the point    along the line

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

Q : 14     In what ratio, the line joining    and    is divided by the line   ?

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

Q : 13     Show that the equation of the line passing through the origin and making an angle
with the line   is    .

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

Q: 12     Find the equation of the line passing through the point of intersection of the lines    and    that has equal intercepts on the axes.

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

Q: 11     Find the equation of the lines through the point which make an angle of   with the line

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,...

Q : 10       If three lines whose equations are   and    are concurrent, then show that

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

Q : 9     Find the value of so that the three lines    and     may intersect at one point.

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

Q : 8     Find the area of the triangle formed by the lines    and

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

Q : 7      Find the equation of a line drawn perpendicular to the line
through the point, where it meets the -axis.

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

Q : 6       Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines    and  .

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

Q : 5     Find perpendicular distance from the origin to the line joining the points    and

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