JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 8)
Let $$\mathrm{A}(-1,1)$$ and $$\mathrm{B}(2,3)$$ be two points and $$\mathrm{P}$$ be a variable point above the line $$\mathrm{AB}$$ such that the area of $$\triangle \mathrm{PAB}$$ is 10. If the locus of $$\mathrm{P}$$ is $$\mathrm{a} x+\mathrm{by}=15$$, then $$5 \mathrm{a}+2 \mathrm{~b}$$ is :
$$-\frac{12}{5}$$
$$-\frac{6}{5}$$
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Explanation
_5th_April_Evening_Shift_en_8_1.png)
$\begin{aligned} & \frac{1}{2}\left|\begin{array}{ccc}h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|=10 \\\\ & -2 x+3 y=25 \\\\ & -\frac{6}{5} x+\frac{9}{5} y=15\end{aligned}$
$\begin{aligned} & a=-\frac{6}{5}, b=\frac{9}{5} \\\\ & 5 a=-6,2 b=\frac{18}{5}\end{aligned}$
$$ \therefore $$ $$5 \mathrm{a}+2 \mathrm{~b}$$ = $$-\frac{12}{5}$$
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