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Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

For a point P...

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

Let $P = (h, k)$ ,
$d_{1} (P) = \frac{| h - k |}{\sqrt{2}} d_{2} (P) = \frac{| h + k |}{\sqrt{2}}$
Now,
$2 \leq d_{1} (P) + d_{2} (P) \leq 4 2 \leq \frac{| h - k |}{\sqrt{2}} + \frac{| h + k |}{\sqrt{2}} \leq 4 2 \sqrt{2} \leq | h - k | + | h + k | \leq 4 \sqrt{2}$

When $h \geq k$
$2 \sqrt{2} \leq h - k + h + k \leq 4 \sqrt{2} 2 \sqrt{2} \leq 2 h \leq 4 \sqrt{2} \sqrt{2} \leq h \leq 2 \sqrt{2}$

Similarly, when $k > h$
$\sqrt{2} \leq k \leq 2 \sqrt{2}$

Area of the region is,
$= 2 \sqrt{2} \times 2 \sqrt{2} - \sqrt{2} \times \sqrt{2} = 8 - 2 = 6$

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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

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