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P1, P2 are points on either of the two lines y – 3x = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P?, P? on the bisector of the angle between the given lines.

(Hint: Lines are y = 3 x + 2 and y = – 3 x + 2 according as x ≥ 0 or x < 0

Y-axis is the bisector of the angles between the lines. P?, P? are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on y-axis as the common foot of perpendiculars from these points. The y-coordinate of the foot of the perpendicular is given by 2 + 5 cos30°.)

0 4 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Sol:

    Given  lines  are  y−3|x|=2⇒                                 y−3x=2,  if  x≥0                               …(i)and                              y+3x=2,  if  x<0                               …(ii)Slope  of  eqn.(i)  is  tanθ=3  ∴θ=600Slope  of  eqn.(ii)  is  tanθ=−3  ∴θ=1200Solving  eqn.(i)  and  eqn.(ii)  we  get                y  −  3x  =  2                y  +  3x  =  2_                             2y=4           ⇒y=2Putting  the  value  of  y  in  eqn.(i)  we  get                x=0∴    of    of  line  (i)  and  (ii)  is  Q  (0,2)∴             QO=2In  ΔPEQ,              cos300=PQQE⇒                 32=PQ5      ⇒PQ=532∴         OP=OQ+PQ                    =2+532Hence,  the  coordinates  of  the  foot  of  perpendicular=(0,2+532)

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