Question

The line 3x-8y=k cuts the x-axis and y- axis at the point A and B respectively. If the area of ∆AOB =12sq.units, find the value of k.​

Answer 1

Given:

The equation is:

`3x-8y=k`

It cuts the x-axis and y- axis at the point A and B respectively.

The area of ∆AOB =12 sq.units.

To find:

The value of k.

Solution:

We have,

`3x-8y=k`

Substituting
`x=0` to find the y-intercept.

`3(0)-8y=k`

`0-8y=k`

`y=(k)/(-8)`

`y=-(k)/(8)`

Substituting
`y=0` to find the x-intercept.

`3x-8(0)=k`

`3x-0=k`

`x=(k)/(3)`

Area of a triangle is:

`A=(1)/(2)* base* height`

The height of the ∆AOB is
`OB=(k)/(8)` because distance cannot be negative and the base of the ∆AOB is
`OA=(k)/(3)`. So, the area of the ∆AOB is:

`A=(1)/(2)* (k)/(8)* (k)/(3)`

`A=(k^2)/(48)`

It is given that, the area of ∆AOB = 12 sq.units.

`(k^2)/(48)=12`

`k^2=576`

`k=\pm √(576)`

`k=\pm 24`

Therefore, the value of k is either 24 or -24.