Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x + 2y = 13.

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)