JEE Mains · Maths · STD 11 - 9. straight line
Two vertices of a triangle \(\mathrm{ABC}\) are \(\mathrm{A}(3,-1)\) and \(\mathrm{B}(-2,3)\), and its orthocentre is \(\mathrm{P}(1,1)\). If the coordinates of the point \(\mathrm{C}\) are \((\alpha, \beta)\) and the centre of the circle circumscribing the triangle \(\mathrm{PAB}\) is \((h, k)\), then the value of \((\alpha+\beta)+2(h+k)\) equals :
- A \(51\)
- B \(81\)
- C \(5\)
- D \(15\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}\) Equation of \(PC\) is \(y-1=\frac{5}{4}(x-1)\)...............(\(1\)) \(\mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1\) Equation of \(\mathrm{BC}\) is…
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