(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.

(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.

(i) Below figure shows the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5:

Given,

⇒ 2y = 3x + 5

⇒ y = $\dfrac{3}{2}x + \dfrac{5}{2}$

Comparing above equation with y = mx + c we get,

m = $\dfrac{3}{2}$

Let slope of line AB be m₁.

Since, AB and line 2y = 3x + 5 are perpendicular.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

$\dfrac{3}{2} \times m_1 = -1$

$m_1 = -\dfrac{2}{3}$.

By point-slope form,

Equation of AB : y - y₁ = m(x - x₁)

⇒ y - 2 = $-\dfrac{2}{3}$(x - 3)

⇒ 3(y - 2) = -2(x - 3)

⇒ 3y - 6 = -2x + 6

⇒ 2x + 3y = 6 + 6

⇒ 2x + 3y = 12.

Hence, equation of AB is 2x + 3y = 12.

(ii) Below figure shows AB with its intercepts on x axis and y axis:

At A,

y co-ordinate = 0 as it lies on x-axis.

Substituting y = 0 in equation of AB,

⇒ 2x + 3(0) = 12

⇒ 2x = 12

⇒ x = 6.

A = (x, 0) = (6, 0).

At B,

x co-ordinate = 0 as it lies on y-axis.

Substituting x = 0 in equation of AB,

⇒ 2(0) + 3y = 12

⇒ 3y = 12

⇒ y = 4.

B = (0, y) = (0, 4).

Area of right angle triangle OAB = $\dfrac{1}{2} \times OA \times OB$

= $\dfrac{1}{2} \times 6 \times 4$

= 12 sq. units.

Hence, A = (6, 0), B = (0, 4) and area of triangle OAB = 12 sq. units.