question_answer

Let the equations of two sides of a triangle be \[3x-2y+6=0\] and\[4x+5y-20=0\]. If the orthocentre of this triangle is at \[\left( 1,\text{ }1 \right)\], then the equation of its third side is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) \[26x\,-\,122y\,-\,1675\,=\,0\]

B)               \[26x\,+61y\,+\,1675\,=\,0\]

C)               \[122y\,-\,\,26x\,-\,1675\,=\,0\]

D)               \[122y\,+\,\,26x\,+\,1675\,=\,0\]

Correct Answer: A

Solution :

                        \[4x+5y-20=0\]             ..... (1) \[3x-2y+6=0\]               ..... (2) orthocentre is (1, 1) line perpendicular to \[4x+5y-20=0\] and passes through (1, 1) is \[\left( y-1 \right)=\frac{5}{4}\left( x-1 \right)\] \[\Rightarrow \,\,\,\,\,5x-4y=1\]    ..... (3) and line \[\bot \]to \[3x-2y+6=0\] and passes through (1, 1) \[y-1=-\frac{2}{3}\left( x-1 \right)\] \[\Rightarrow \,\,\,\,\,2x+3y=5\]               ..... (4) Solving (1) and (4) we get C \[\left( \frac{35}{2},\,\,-10 \right)\] Solving (2) and (3) we get A\[\left( -13,\,\,\,\,\frac{-33}{2}, \right)\] Side BC is \[y+10=\frac{\frac{-33}{2}+10}{-13\,-\,\frac{35}{2}}\left( x-\frac{35}{2} \right)\] \[\Rightarrow \,\,\,\,\,y+10\,\,=\,\,\frac{13}{61}\,\left( x-\frac{35}{2} \right)\] \[\Rightarrow \,\,\,\,26x-122y-1675=0\]