AP EAMCET · Maths · Three Dimensional Geometry
If \(\mathrm{A}(0,0,0), \mathrm{B}(3,4,0), \mathrm{C}(0,12,5)\) are the vertices of a triangle ABC, then the \(x\)-coordinate of its incentre is
- A \(\frac{25}{18+7 \sqrt{2}}\)
- B \(\frac{25}{26}\)
- C \(\frac{39}{18+7 \sqrt{2}}\)
- D \(\frac{39}{26}\)
Answer & Solution
Correct Answer
(C) \(\frac{39}{18+7 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(a = \sqrt{(0-3)^2 + (12-4)^2 + (5-0)^2} = \sqrt{9+64+25} = \sqrt{98} = 7\sqrt{2}\) \(b = \sqrt{(0-0)^2 + (12-0)^2 + (5-0)^2} = \sqrt{0+144+25} = \sqrt{169} = 13\) \(c = \sqrt{(3-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{9+16+0} = \sqrt{25} = 5\)…
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