Question

If the plane $\frac{x}{3} + \frac{y}{2} - \frac{z}{4} = 1$ cuts the co-ordinate axes at points A, B and C, then the area of the triangle ABC is

• A. $\frac{\sqrt{61}}{2}$ sq. units
• B. $2\sqrt{61}$ sq. units
• C. $\sqrt{61}$ sq. units
• D. $3\sqrt{61}$ sq. units

Hint:

Triangle area between intercepts $a, b, c$ is $\frac{1}{2}\sqrt{a^2b^2 + b^2c^2 + c^2a^2}$.

Answer 1

The Correct Option is C

Step 1: Concept
The intercepts are $a=3, b=2, c=-4$. Points are $A(3,0,0), B(0,2,0), C(0,0,-4)$.

Step 2: Meaning
The area $S$ of a triangle with vertices on the axes is $S = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$.

Step 3: Analysis
$ab = 6$, $bc = -8$, $ca = -12$. Area $= \frac{1}{2} \sqrt{6^2 + (-8)^2 + (-12)^2} = \frac{1}{2} \sqrt{36 + 64 + 144}$. Area $= \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 \times 61} = \frac{1}{2} \cdot 2\sqrt{61}$.

Step 4: Conclusion
Area $= \sqrt{61}$ sq. units. Final Answer: (C)