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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

For a point ...

Question

For a point $P$ in the plane, let $d_{1}$ (P) and $d_{2}$ (P) be the distances of the point P from the lines $x - y = 0$ and $x + y = 0$ respectively. The area of the region R consisting of all points P lying in the first quadrant of the plane and satisfying $2 \leq d_{1} (P) + d_{2} (P) \leq 4$ , is

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Solution

The correct option is B $6$
$2 \leq d_{1} (p) + d_{2} (p) \leq 4$

For $P (α, β), α > β$

$\Rightarrow 2 \sqrt{2} \leq 2 α \leq 4 \sqrt{2}$

$\sqrt{2} \leq α \leq 2 \sqrt{2}$

$\Rightarrow$ Area of region $= ((2 \sqrt{2})^{2} - (\sqrt{2})^{2})$

$= 8 - 2 = 6$ sq. units

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Similar questions

Q. For a point P in the plane, let

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For a point P in the plane, let d1 (P) and d2 (P) be the distances of the point P from the lines x−y=0 and x+y=0 respectively. The area of the region R consisting of all points P lying in the first quadrant of the plane and satisfying 2≤d1(P)+d2(P)≤4, is

The correct option is B 62≤d1(p)+d2(p)≤4For P(α,β),α>β⇒2√2≤2α≤4√2√2≤α≤2√2⇒ Area of region =((2√2)2−(√2)2) =8−2=6 sq. units

The correct option is B $6$
$2 \leq d_{1} (p) + d_{2} (p) \leq 4$

For $P (α, β), α > β$

$\Rightarrow 2 \sqrt{2} \leq 2 α \leq 4 \sqrt{2}$

$\sqrt{2} \leq α \leq 2 \sqrt{2}$

$\Rightarrow$ Area of region $= ((2 \sqrt{2})^{2} - (\sqrt{2})^{2})$

$= 8 - 2 = 6$ sq. units