If the area of the triangle formed by the straight lines x = 0, y = 0 and 3x + 4y = a (a > 0) is 6. Find the value of ‘a’.

The area of triangle formed by the st.line ax + by + c = 0 with the co-ordinates axes is $\frac{{\mathrm{c}}^2}{2\left|\mathrm{ab}\right|}\text{ sq.units }$
Give st.lines x = 0, y = 0 and 3x + 4y – a = 0
$\therefore$The area of the triangle formed by the given st.lines is 6
$\begin{array}{l}\Rightarrow \frac{{\mathrm{c}}^2}{2\left|\mathrm{ab}\right|}=6\Rightarrow \frac{{\mathrm{a}}^2}{2|4.3|}=6\Rightarrow {\mathrm{a}}^2=144\\ \therefore \mathrm{a}=\pm 12\\ \therefore \mathrm{a}=12(\therefore \mathrm{a}>0)\end{array}$