Question

The area of the triangle formed by the plane $2x+3y+6z+9=0$ with $Y-$axis and $Z-$axis (in $\text{sq.units}$) is

• A. $\frac{9}{2}$
• B. $\frac{11}{4}$
• C. $\frac{9}{4}$
• D. $\frac{11}{2}$

Answer 1

The correct option is C $\frac{9}{4}$

Given :
$2x+3y+6z+9=0$
converting into intercept form :
$\Rightarrow \frac{x}{-9/2}+\frac{y}{-3}+\frac{z}{-3/2}=1$
So, intercepts on $X,Y,Z$ axes are $a=-\frac{9}{2},b=-3,c=-\frac{3}{2}$ respectively.
So, required area of triangle $=\frac{1}{2}|bc|$
$=\frac{1}{2}\left|\right(-3\left)(-\frac{3}{2})\right|=\frac{9}{4}\text{sq.units}$