Answer :

Here we have a perpendicular from the origin i.e.(0,0) to the


straight line 3x + 2y = 13.


We have to find the foot of the perpendicular i.e the intersection point at the line 3x + 2y = 13.


As these lines are perpendicular the product of their slopes is equal to –1.


Slope of the 3x + 2y–13 = 0 is m.



Therefore the slope of perpendicular is


i.e.


Hence the equation of the perpendicular from (0,0) and slope as


is (y–y1) = m(x–x1)




3y = 2x


3y–2x = 0


2x–3y = 0


Now solve the two equations 3x + 2y–13 = 0 and 2x–3y = 0.


3x + 2y–13 = 0–––(1)


2x–3y = 0–––––(2)


Multiply (1) by 3 and (2) by 2 and add




Substitute x = 3 in the equation 2x–3y = 0.


2(3)–3y = 0


–3y = –6


y = 2(Divide both the sides of the equation by –3)


Hence the coordinates of the foot of the perpendicular is(3,2)


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