QJEEM19Mae848: JEE Main 2019 Mathematics - Class 11 JEEM19Mae848 Solution: Let the equations of two sides of a triangle be 3x 2y + 6 = 0…

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

(y $-$ 1) = ${5 \over 4}$ (x $-$ 1)

$\Rightarrow$   5x $-$ 4y = 1       . . .(3)

and line $\bot$ to 3x $-$ 2y + 6 = 0

and passes through (1, 1)

y $-$ 1 = $-$ ${2 \over 3}$ (x $-$ 1)

$\Rightarrow$   2x + 3y = 5       . . .(4)

Solving (1) and (4) we get C $\left( {{{35} \over 2}, - 10} \right)$

Solving (2) and (3) we get A $\left( { - 13,{{ - 33} \over 2}} \right)$

Side BC is y + 10 = ${{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$   y + 10 = ${{13} \over {61}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$   26x $-$ 122y $-$ 1675 = 0