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The equation of the given line is
3
x+y+2=0
this equation can be reduced as −
3
x−y=2
on dividing both sides by
(−
3
)
2
+(−1)
2
=2,
we obtain −
2
3
x−
2
1
y=
2
2
⇒{−
2
3
}x+{−
2
1
}y=1....(1)
On comparing equation (1) to xcosθ+ysinθ=p,
we obtain cosθ=−
2
3
,sinθ=−
2
1
, and p=1
Since the value of sinθ and cosθ are both negative, θ is in the third quadrant
∴θ=π+
6
π
=
6
7π
Thus, the respective values of θ and p are
6
7π
and 1
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