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
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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