26. Find the equations of the straight lines
which make 150° angle with the
positive X-axis in the positive direction and
which pass through the points (-2,-1)
Answers
Answer :
x + √3y + 2 + √3 = 0
Note :
The equation of the straight line passing through the point (x1 , y1) and making an angel θ with the +ve x-axis in positive direction (anti-clockwise direction) is given by ;
(x - x1)•tanθ = (y - y1)
Solution :
Here ,
It is given that , the required straight line is making an angel of 150° with the +ve x-axis in positive direction (anti-clockwise direction) .
Thus ,
θ = 150°
Also ,
It is given that , the required line is passing through the point (-2 , -1) .
Thus ,
(x1 , y1) = (-2 , -1)
ie. x1 = -2 , y1 = -1
Now ,
The equation of the required line will be given as ;
=> (x - x1)•tanθ = (y - y1)
=> [x - (-2)]•tan150° = [y - (-1)]
=> (x + 2)•tan(180° - 30°) = y + 1
=> (x + 2)•(-tan30°) = y + 1
=> (x + 2)•(-1/√3) = y + 1
=> x + 2 = -√3•(y + 1)
=> x + 2 = -√3y - √3 = 0
=> x + √3y + 2 + √3 = 0
Hence ,
The required equation of line is :
x + √3y + 2 + √3 = 0
Answer:
Equation of any line is given by y=mx+c
where m=tanθ, θ is angle made by x− axis
m=tan150
o
=tan(180
o
−30
o
)
=−tan30
o
=−
3
1
also when x=0, we have y=2.
so y=
3
−x
+c put x=0 and y=2
2=c
so equation is y=
3
−x
+2⇒x+y
3
=2
3