Straight Lines
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A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O n the ground is 45o. It flies off horizontally straight away from the point O. After 1s, the elevation of the bird from O is reduced to 30o. Then, the speed (in m/s of the bird is
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(
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-
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D.
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A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2 , then the height above the point P from where the body began to fall is
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720 m
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900 m
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320 m
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680 m
A.
720 m
Subtracting we get 400 = 8g + 4gt
⇒ t = 8 sec
∴
∴ Desired height = 320 + 400 = 720 m.
A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is
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-1/4
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-4
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-2
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-1/2
C.
-2
The slope of line PQ 
Let 'm' be the slope of the line PQ, then the equation of PQ is
y -2 = m (x-1)
Now, PQ meets X-axis at P 
and y-axis at Q (0,2-m)
⇒ 
Now, f'(m) = 0
m = ± 2
f(2) =0
f(-2) = 8
Since, the area cannot be zero, hence the required value of m is -2
A ray of light along 
get reflected upon reaching X -axis, the equation of the reflected ray is
B.
Given equation of line
Slope of incident ray is 
So, slope of reflected ray must be 
and the point of incident 
So equation of reflected ray
A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is
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x + y = 7
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3x − 4y + 7 = 0
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4x + 3y = 24
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3x + 4y = 25
C.
4x + 3y = 24
The equation of axes is xy = 0
⇒ the equation of the line is
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