JEE Mains · Maths · STD 11 - Trigonometrical equations
A triangle \(ABC\) lying in the first quadrant has two vertices as \(A (1,2)\) and \(B (3,1)\). If \(\angle BAC =90^{\circ},\) and \(\operatorname{ar}(\Delta ABC )=5 \sqrt{5}\) sq. units then the abscissa of the vertex \(C\) is
- A \(2+\sqrt{5}\)
- B \(1+\sqrt{5}\)
- C \(1+2 \sqrt{5}\)
- D \(2 \sqrt{5}-1\)
Answer & Solution
Correct Answer
(C) \(1+2 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{ K -2}{ h -1}\right)\left(\frac{1-2}{3-1}\right)=-1 \Rightarrow K =2 h\) \(\sqrt{5}| h -1|=10\) \(\because[\Delta ABC ]=5 \sqrt{5}\) \(\Rightarrow \frac{1}{2}(\sqrt{5}) \sqrt{( h -1)^{2}+( K -2)^{2}}=5 \sqrt{5}\) \(\Rightarrow h =2 \sqrt{5}+1( h >0)\)
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