106Q8. answer is (b)
see. y=tan120°x + c
==>y=-√3x + c
==>√3x + y + c = 0
Dist. from origin is 5
==> lcl/√(√3)2 + 12 = 5
==> lcl = 10
==> c = ±10
So, √3x+ y ±10 = 0
Clearly (b) is one of the answer
check these ques out from t 26
Q8) Find the eqn. of the line which is at distance 5 from origin it makes an angle of 120o with +ve x-axis.
a) x+√3y-10=0
b) √3x+y-10=0
c) √3x+y-5=0
d) √3x-y-10=0
Q2) The two points of a triangle are (4,3) and (9,8). The third point lies on the line x-y+3=0. Find the area.
a) 20
b) 10
c) 0
d) 10√2
Q10 Find the image of the point (2,2) in the line mirror 3x+4y-5=0.
a) (-2/25,-1)
b) (-4/25,22/25)
c) (-1/25,-26/25)
d) none
the ans of 10th ques is given to be b but check whether it lies o the perpendicular??????????/
the ans of 2nd ques is 10 but is mein kis ka area find karna hai us line pe infinite points hain?????????
8th ques mein is mein 5 perp dist hai x intercept hai ya y intercept hai???????????
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5 Answers
106
62Q9 the line is perpendicular to the base.
area is 1/2. b . h
base is the distance between the two points
Height is the distance of the pont from the line.
So for your question, the position fo the point on the line wont matter.
11sir 2nd ques mein line formed by 2 points is parallel to the given line!
11ek aur ques from t12
Q8 minimize x4+ 1/x4 +x2+ 1/x2+ x+ 1/x+ x3+ 1/x3 +1
a) 0
b) 1
c) -3
d) 9
Your Answer: (b)
Correct Answer: (d)
Score: -1
maine x=-1 dala tha