Question

The line 3x − 4y − 12 = 0 meets x-axis at point A and y-axis at point B.

1. Find the co-ordinates of the point P on line segment AB dividing AB in the ratio 2 : 1.
2. Find the equation of the line that passes through the point P and is perpendicular to AB.

Answer 1

(i) Given,

The line 3x − 4y − 12 = 0 meets x-axis at point A and y-axis at point B.

Substituting x = 0, we get:

⇒ 3.0 − 4y − 12 = 0

⇒ 4y = −12

⇒ y = −3

B = (0, −3)

Substituting y = 0, we get:

⇒ 3x − 4.0 − 12 = 0

⇒ 3x = 12

⇒ x = `12/3​` = 4

A = (4, 0)

By section formula,

Point of division = `((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))`

Substituting values we get:

⇒ P = `((2×0+1×4)/(2+1)​, (2×−3+1×0)/(2+1​))`

⇒ P = `(4/3, (-6)/3)`

⇒ P = `(4/3, -2)`

Hence, coordinates of P = `(4/3, -2)`

(ii) Slope of AB = `(−3−0)/(0−4)`

= `(−3)/(−4)`

= `3/4`

We know that,

Product of slope of perpendicular lines = −1. Let slope of line perpendicular to AB be m.

⇒ `m × 3/4​ = −1` 

⇒ `m = −4/3​`

By point slope form:

Equation of line:

y − y₁ = m(x − x₁)

Equation of line passing through P and perpendicular to AB is:

⇒ `y−(−2) = −4/3​(x − 4/3​)`

⇒ `y + 2 = −4/3​(x − 4/3​)`

⇒ `3(y + 2) = −4(x − 4/3​)`

⇒ `3y + 6 = −4x + 16/3​`

⇒ 3(3y + 6) = −12x + 16

⇒ 9y + 18 = −12x + 16

⇒ 12x + 9y + 2 = 0

Hence, equation of line passing through P and perpendicular to AB is 12x + 9y + 2 = 0.