Question

The area of the triangle formed by plane $2x+3y+6z+9=0$ with $y$ - axis, $z$- axis is (in Sq.units)

• A. $9$
• B. $\frac{9}{2}$
• C. $\frac{9}{4}$
• D. $\frac{9}{8}$

Answer 1

Answer: (C)

$\frac{9}{4}$

Given equation of plane is $2x+3y+6z+9=0$.
So, intercept made by this plane in $y$ and $z$ axis are, $\frac{-3}{2},-3$.
Hence, area of triangle made by this intercept and $y$ and $z$ axis is
$\frac{1}{2}\cdot \mid \frac{-3}{2}\mid \cdot \mid -3\mid =\frac{9}{4}$ square unit