JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 8)
A line passing through the point $$\mathrm{A}(9,0)$$ makes an angle of $$30^{\circ}$$ with the positive direction of $$x$$-axis. If this line is rotated about A through an angle of $$15^{\circ}$$ in the clockwise direction, then its equation in the new position is :
$$\frac{y}{\sqrt{3}+2}+x=9$$
$$\frac{x}{\sqrt{3}+2}+y=9$$
$$\frac{x}{\sqrt{3}-2}+y=9$$
$$\frac{y}{\sqrt{3}-2}+x=9$$
Explanation
_30th_January_Morning_Shift_en_8_1.png)
$$\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)$$
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