Question

If $3x–4y+12=0$ and $3x+2y–6=0$, then area of triangle formed by the given lines and x-axis is (in sq. units)

• A. 8
• B. 9
• C. 11
• D. 12

Answer 1

The correct option is B 9

For the line $3x–4y+12=0$:
At $x=0;3\left(0\right)–4y+12=0$
$\Rightarrow –4y=–12$
$\Rightarrow y=3$
At $y=0;3x–4\left(0\right)+12=0$
$\Rightarrow 3x=–12$
$\Rightarrow x=–4$

$\begin{array}{ccc}x& 0& -4\\ y& 3& 0\end{array}$

Now, for the line $3x+2y–6=0$:
At $x=0,3\left(0\right)+2y–6=0$
$\Rightarrow 2y=6$
$\Rightarrow y=3$
At $y=0;3x+2\left(0\right)–6=0$
$\Rightarrow 3x=6$
$\Rightarrow x=2$
$\begin{array}{ccc}x& 0& 2\\ y& 3& 0\end{array}$

Drawing the lines $3x–4y+12=0$ and $3x+2y–6=0$ and x-axis on a coordinate plane, we get


These lines form a triangle such that its height is 3 units and base length is 6 units.
$\therefore$ Area of Triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 6\times 3$
= 9 sq.units
Hence, the correct answer is option (b).