Question

(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 y-axis at B. Calculate the area of triangle OAB, where O is the origin.

Answer 1

(i) 2y = 3x + 5

Slope of this line =

Slope of the line AB =

(x₁, y₁) = (3, 2)

The required equation of the line AB is

y - y₁ = m(x - x₁)

y - 2 = (x - 3)

3y - 6 = -2x + 6

2x + 3y = 12

(ii) For the point A (the point on x-axis), the value of y = 0.

2x + 3y = 12 2x = 12 x = 6

Co-ordinates of point A are (6, 0).

For the point B (the point on y-axis), the value of x = 0.

2x + 3y = 12 3y = 12 y = 4

Co-ordinates of point B are (0, 4).

Area of OAB = OA $\times$ OB = 6 $\times$ 4 = 12 sq units