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Question
If the plane \[\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1\] cuts the coordinate axes at points A, B and C, then the area of the X triangle ABC is
Options
\[\frac{\sqrt{61}}{2}\] sq. units
2\[{\sqrt{61}}{}\] sq. units
\[{\sqrt{61}}{}\] sq. units
3\[{\sqrt{61}}{}\]sq. units
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Solution
\[{\sqrt{61}}{}\] sq. units
Explanation:
For A, y = 0, z = 0
∴ A ≡ (3, 0, 0)
For B, x = 0, z = 0
∴ B ≡ (0, 2, 0)
For C, x = 0, y = 0
∴ C ≡ (0, 0, −4)
If a triangle cuts intercepts a, b, c on X, Y and Z axes, then the area of the triangle is given by
\[\mathrm{A}=\frac{1}{2}\sqrt{\mathrm{a}^{2}\mathrm{b}^{2}+\mathrm{b}^{2}\mathrm{c}^{2}+\mathrm{a}^{2}\mathrm{c}^{2}}\]
\[\therefore\quad\text{Required area}=\frac{1}{2}\sqrt{9\times4+4\times16+16\times9}\]
\[=\frac{1}{2}\sqrt{36+64+144}\]
\[=\frac{1}{2}\sqrt{244}\]
\[=\frac{1}{2}\times2\sqrt{61}\]
\[{\sqrt{61}}{}\] sq. units
