QJEEM22Ma0fea: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0fea Solution: The equations of the sides and of a triangle are and…

Slope of $AH = {{a + 2} \over 1}$

Slope of $BC = - {1 \over p}$

$\therefore$ $p = a + 2$ ...... (i)

Coordinate of $C = \left( {{{18p - 30} \over {p + 1}},\,{{15p - 33} \over {p + 1}}} \right)$

Slope of $HC = {{{{15P - 33} \over {p + 1}} - a} \over {{{18p - 30} \over {p + 1}} - 2}} = {{15p - 33 - (p - 2)(p + 1)} \over {18p - 30 - 2p - 2}}$

$= {{16p - {p^2} - 31} \over {16p - 32}}$

$\because$ ${{16p - {p^2} - 31} \over {16p - 32}} \times - 2 = - 1$

$\therefore$ ${p^2} - 8p + 15 = 0$

$\therefore$ $p = 3$ or $5$

But if $p = 5$ then $a = 3$ not acceptable

$\therefore$ $p = 3$