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Question: If a point P moves such that its distance from line y = \(\sqrt{3}\)x –7 is same as its distance fro...

If a point P moves such that its distance from line y = 3\sqrt{3}x –7 is same as its distance from (23\sqrt{3}, –1), then area of curve described by P, enclosed between coordinate axes is-

A

32\frac{\sqrt{3}}{2}

B

23\sqrt{3}

C

6

D

None of these

Answer

32\frac{\sqrt{3}}{2}

Explanation

Solution

The point (23\sqrt{3}, –1) lie on line y =3\sqrt{3}x –7

\ locus of point is straight line perpendicular to

given line passing through given point

i.e. x + y3\frac{y}{\sqrt{3}}= 1

\ Area enclosed = 3×12\frac{\sqrt{3} \times 1}{2}= 32\frac{\sqrt{3}}{2}