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Question: If we reduce \(3 x + 3 y + 7 = 0\) to the form \(x \cos \alpha + y \sin \alpha = p\), then the value...

If we reduce 3x+3y+7=03 x + 3 y + 7 = 0 to the form xcosα+ysinα=px \cos \alpha + y \sin \alpha = p, then the value of p is

A

723\frac { 7 } { 2 \sqrt { 3 } }

B

73\frac { 7 } { 3 }

C

372\frac { 3 \sqrt { 7 } } { 2 }

D

732\frac { 7 } { 3 \sqrt { 2 } }

Answer

732\frac { 7 } { 3 \sqrt { 2 } }

Explanation

Solution

Given equation is 3x+3y+7=03 x + 3 y + 7 = 0, Dividing both sides by 32+32\sqrt { 3 ^ { 2 } + 3 ^ { 2 } }

3x32+32+3y32+32+732+32=0\frac { 3 x } { \sqrt { 3 ^ { 2 } + 3 ^ { 2 } } } + \frac { 3 y } { \sqrt { 3 ^ { 2 } + 3 ^ { 2 } } } + \frac { 7 } { \sqrt { 3 ^ { 2 } + 3 ^ { 2 } } } = 0

332x+332y=732\frac { 3 } { 3 \sqrt { 2 } } x + \frac { 3 } { 3 \sqrt { 2 } } y = \frac { - 7 } { 3 \sqrt { 2 } } , \therefore p=732=732p = \left| \frac { - 7 } { 3 \sqrt { 2 } } \right| = \frac { 7 } { 3 \sqrt { 2 } }.