JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 19)
The equations of the sides $$\mathrm{AB}, \mathrm{BC}$$ and $$\mathrm{CA}$$ of a triangle $$\mathrm{ABC}$$ are $$2 x+y=0, x+\mathrm{p} y=15 \mathrm{a}$$ and $$x-y=3$$ respectively. If its orthocentre is $$(2, a),-\frac{1}{2}<\mathrm{a}<2$$, then $$\mathrm{p}$$ is equal to ______________.
Answer
3
Explanation
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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
$$\therefore$$ $$p = 3$$
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