Question

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

Answer 1

i. 2y = 3x + 5

`=> y = (3x)/2 + 5/2`

Slope of this line = `3/2`

Slope of the line AB = `(-1)/(3/2) = (-2)/3`

(x₁, y₁) = (3, 2)

The required equation of the line AB is

y − y₁ = m(x − x₁)

`y - 2 = (-2)/3 (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 ΔOAB = `1/2 xx OA xx OB`

= `1/2 xx 6 xx 4`

= 12 sq units