Question

The plane $\frac{x}{y}+\frac{y}{3}+\frac{z}{4}$ =1 cutes the axes in A,B,C, then the are of the $\Delta ABC$ is;

• A. $\sqrt{29}$
• B. $\sqrt{41}$
• C. $\sqrt{61}$
• D. None of these

Answer 1

The correct option is C $\sqrt{61}$

We have,

$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$

Compare this equation of plane,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Then,

$a=2,b=3,c=4$

Then,

The coordinate of X-axis is $=A(a,0,0)$$=A(2,0,0)$

The coordinate of Y-axis is $=B(0,b,0)$$=A(0,3,0)$

The coordinate of X-axis is $=C(0,0,c)$$=C(0,0,2)$

We know that, the area of $$\Delta ABC$$ is

$Areaof\Delta ABC=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$

$=\frac{1}{2}\sqrt{2^2\times 3^2+3^2\times 4^2+4^2\times 2^2}$

$=\frac{1}{2}\sqrt{36+144+64}$

$=\frac{1}{2}\sqrt{180+64}$

$=\frac{1}{2}\sqrt{244}$

$=\frac{1}{2}\sqrt{2\times 2\times 61}$

$=\sqrt{61}$

Hence, this is the answer.