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.

$3(0)-8y=k$

$0-8y=k$

$y=\dfrac{k}{-8}$

$y=-\dfrac{k}{8}$

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

$3x-8(0)=k$

$3x-0=k$

$x=\dfrac{k}{3}$

Area of a triangle is:

$A=\dfrac{1}{2}\times base\times height$

The height of the ∆AOB is $OB=\dfrac{k}{8}$ because distance cannot be negative and the base of the ∆AOB is $OA=\dfrac{k}{3}$. So, the area of the ∆AOB is:

$A=\dfrac{1}{2}\times \dfrac{k}{8}\times \dfrac{k}{3}$

$A=\dfrac{k^2}{48}$

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

$\dfrac{k^2}{48}=12$

$k^2=576$

$k=\pm \sqrt{576}$

$k=\pm 24$

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