Straight Lines

More Topic from Mathematics

Question 1

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

  • (40 square root of 2 minus end root space 1

  • 40 left parenthesis square root of 3 space space end root space minus space square root of 2 right parenthesis
  • 20 space square root of 2
  • 20 left parenthesis square root of 3 minus 1 end root right parenthesis

Solution

D.

20 left parenthesis square root of 3 minus 1 end root right parenthesis
Question 2

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

  • 720 m

  • 900 m

  • 320 m

  • 680 m

Solution

A.

720 m


straight h space equals space 1 half gt squared space and space straight h space plus space 400 space equals space 1 half space straight g space left parenthesis straight t plus 4 right parenthesis squared
Subtracting we get 400 = 8g + 4gt
⇒ t = 8 sec
∴ straight h space equals space 1 half space straight x space 10 space straight x space 64 space equals space 320 space straight m
∴ Desired height = 320 + 400 = 720 m.
Question 3

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

  • -1/4

  • -4

  • -2

  • -1/2

Solution

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 open parentheses 1 minus 2 over straight m comma space 0 close parentheses and y-axis at Q (0,2-m)
OP space equals space 1 minus 2 over straight m space and space OQ space equals space 2 minus straight m
Also comma space area space space of space increment OPQ space equals space 1 half space left parenthesis OP right parenthesis left parenthesis OQ right parenthesis
space equals space 1 half open vertical bar open parentheses 1 minus 2 over straight m close parentheses left parenthesis 2 minus straight m right parenthesis close vertical bar
equals space 1 half open vertical bar 2 minus straight m minus 4 over straight m plus 2 close vertical bar
equals space 1 half open vertical bar 4 minus open parentheses straight m plus 4 over straight m close parentheses close vertical bar
Let space straight f left parenthesis straight m right parenthesis space equals space 4 minus open parentheses straight m plus 4 over straight m close parentheses
rightwards double arrow space straight f apostrophe left parenthesis straight m right parenthesis space equals space minus space 1 plus 4 over straight m squared
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

Question 4

A ray of light along straight x space plus square root of 3 straight y end root space equals space square root of 3 get reflected upon reaching X -axis, the equation of the reflected ray is 

  • straight y equals space straight x plus square root of 3
  • square root of 3 straight y end root space equals space straight x minus square root of 3
  • straight y space equals space square root of 3 straight x end root minus square root of 3
  • square root of 3 straight y end root space equals space straight x minus 1

Solution

B.

square root of 3 straight y end root space equals space straight x minus square root of 3

Given equation of line
straight x plus square root of 3 straight y space equals square root of 3 space end root space space space space space.... space left parenthesis straight i right parenthesis
straight y equals space 1 minus fraction numerator straight x over denominator square root of 3 end fraction
Slope of incident ray is space minus fraction numerator 1 over denominator square root of 3 end fraction 
So, slope of reflected ray must be fraction numerator 1 over denominator square root of 3 end fraction and the point of incident left parenthesis square root of 3 comma 0 right parenthesis
So equation of reflected ray
straight y minus 0 space equals space fraction numerator 1 over denominator square root of 3 end fraction space left parenthesis straight x minus square root of 3 right parenthesis
rightwards double arrow space square root of 3 straight y end root space equals space straight x minus square root of 3

Question 5

A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is

  • x + y = 7

  • 3x − 4y + 7 = 0

  • 4x + 3y = 24

  • 3x + 4y = 25

Solution

C.

4x + 3y = 24

The equation of axes is xy = 0 
⇒ the equation of the line is
fraction numerator straight x.4 space plus straight y.3 over denominator 2 end fraction space equals space 12
rightwards double arrow space 4 straight x space plus 3 straight y space equals space 24