JEE Mains · Maths · STD 11 - 9. straight line
Let the equation of two sides of a triangle be \(3x\,-\,2y\,+\,6\,=\,0\) and \(4x\,+\,5y\,-\,20\,=\,0.\) If the orthocentre of this triangle is at \((1, 1),\) then the equation of its third side is
- A \(122y\, - \,26x\, - 1675\, = \,0\)
- B \(26x\, + \,61y\, + \,1675\, = \,0\)
- C \(122y\, + \,26x\, + 1675\, = \,0\)
- D \(26x\, - \,122y\, - \,1675\, = \,0\)
Answer & Solution
Correct Answer
(D) \(26x\, - \,122y\, - \,1675\, = \,0\)
Step-by-step Solution
Detailed explanation
Equation of \(AB\) is \(3x-2y+6=0\) Equation of \(AC\) is \(4x+5y-20=0\). Equation of \(BE\) is \(2x+3y-5=0\) Equation of \(CF\) is \(5x-4y-1=0\) \( \Rightarrow \) Equation of \(BC\) is \(26x-122y=1675\)
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