JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 12)
Let A be the set of all points ($$\alpha$$, $$\beta$$) such that the area of triangle formed by the points (5, 6), (3, 2) and ($$\alpha$$, $$\beta$$) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :
$${4 \over {\sqrt 5 }}$$
$${16 \over {\sqrt 5 }}$$
$${8 \over {\sqrt 5 }}$$
$${12 \over {\sqrt 5 }}$$
Explanation
_31st_August_Evening_Shift_en_12_1.png)
$$\left| {{1 \over 2}\left| {\matrix{ 5 & 6 & 1 \cr 3 & 2 & 1 \cr \alpha & \beta & 1 \cr } } \right|} \right| = 12$$
4$$\alpha$$ $$-$$ 2$$\beta$$ = $$\pm$$ 24 + 8
$$\Rightarrow$$ 4$$\alpha$$ $$-$$ 2$$\beta$$ = + 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ $$-$$ $$\beta$$ = 16
2x $$-$$ y $$-$$ 16 = 0 ..... (1)
$$\Rightarrow$$ 4$$\alpha$$ $$-$$ 2$$\beta$$ = $$-$$ 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ $$-$$ $$\beta$$ = $$-$$8
2x $$-$$ y + 8 = 0 ...... (2)
Perpendicular distance of (1) from (0, 0)
$$\left| {{{0 - 0 - 16} \over {\sqrt 5 }}} \right| = {{16} \over {\sqrt 5 }}$$
Perpendicular distance of (2) from (0, 0)
$$\left| {{{0 - 0 + 8} \over {\sqrt 5 }}} \right| = {8 \over {\sqrt 5 }}$$
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