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Montano1993 [528]
3 years ago
9

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.​
Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0

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 <em>k</em>.

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.

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