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Practice question on Straight lines for IITJEE , NDA and Airforce 31-45

Straight lines

Question no. 31
The point whose abscissa is equal to its ordinate and which is equidistant from the points (1,0) and (0,3) is
looks_one (1,1)
looks_two (2,2)
looks_3 (3,3)
looks_4 (4,4)
Option looks_two (2,2)
Solution :
point of abscissa (x) = point on ordinate (y) = a

PA = PB
[(a-1)2 + a2]1/2 = [a2 + (a-3)2]1/2
(a-1)2 + a2 = a2 + (a-3)2
a2 + 1 - 2a + a2 = a2 + a2 + 9 - 6a
4a = 8
a = 2
point is (2,2)

Question no. 32
Equation of a line passing through (1,-2) and perpendicular to the line 3x-5y +7 =0 is
looks_one 5x + 3y +1=0
looks_two 3x + 5y +1=0
looks_3 5x - 3y - 1=0
looks_4 3x - 5y + 1=0
Option looks_one 5x + 3y +1=0
Solution :
Slope of line :
3x-5y +7 =0
y = (3/5)x + (7/5)
m1 = 3/5
m1.m2 = -1
m2 = -5/3
y = mx + c
point (1 , -2) satisfy the equation
-2 = (-5/3)(1) + c
c = -1/3
y = (-5/3)x - 1/3
3y = -5x -1

Question no. 33
The line passing through (-1, π/2) and prependicular to √3sinθ + 2cosθ = 4/r is
looks_one 2= √3rsinθ - 2rcosθ
looks_two 5= -2√3rsinθ + 4rcosθ
looks_3 2= √3rsinθ + 2rcosθ
looks_4 5= 2√3rsinθ + 4rcosθ
Option looks_one 2= √3rsinθ - 2rcosθ
Solution :
The points are given in polar form :
acosθ + bsinθ = C/r
differentiate the given equation
√3sinθ + 2cosθ = 4/r
√3cosθ - 2sinθ = 0
√3/2 = tanθ
m1m2 = -1
m2 = -2/√3
√3sinθ - 2cosθ = C/r
Put (-1, π/2) in above equation
-2 = C/-1
C = 2
2= √3rsinθ - 2rcosθ

Question no. 34
The equation of the straight line which is perpendicular to y=x and passes through (3,2) is
looks_one x-y=5
looks_two x+y=5
looks_3 x+y=1
looks_4 x-y=1
Option looks_two x+y=5
Solution :
y = x
Slope of given line (m1) = 1
m1m2 = -1
m2 = -1
y = mx + c
3 = -1(2) + c
c = 5
y = -x + 5
y+x = 5

Question no. 35
A Line AB makes zero intercepts on x - axis and y-axis and it is perpendicular to another line CD, 3x+4y +6= 0. The equation of Line AB is
looks_one y = 4
looks_two 4x - 3y +8=0
looks_3 4x - 3y=0
looks_4 4x - 3y + 6 =0
Option looks_3 4x - 3y=0
Solution :
Line AB makes zero intercept , it passes through origin , c = 0
y = mx
3x + 4y + 6 = 0
m = -3/4
m2 = 4/3
y = 4x/3
3y = 4x
4x -3y = 0

Question no. 36
The equation of line perpendicular to x= c is
looks_one y=d
looks_two x=d
looks_3 x=0
looks_4 None of these
Option looks_one y=d
Solution :
The line prependicular to x =c is y = d

Question no. 37
If we reduce 3x + 3y +7 =0 to the form xcosθ + ysinθ = p , then the value of p is
looks_one
looks_two
looks_3
looks_4
Option looks_4
Solution :
3x+ 3y = -7
x + y = -7/3

Prependicular distance between from origin , P = [(7/6)2 + (7/6)2 ]1/2
P =

Question no. 38
The equation of straight line passing through the intersection of lines x - 2y = 1 and x+ 3y =2 and parallel to 3x + 4y=0 is
looks_one 3x+4y+5=0
looks_two 3x + 4y -10=0
looks_3 3x + 4y - 5=0
looks_4 3x + 4y + 6=0
Option looks_3 3x + 4y - 5=0
Solution :
x - 2y = 1 ......[1]
x + 3y = 2 ......[2]
Solve [1] and [2]
y = 1/5 , x = 7/5
the line will pass through (7/5 , 1/5 )
Slope of line : 3x + 4y = 0
y = -3x/4
m = -3/4
Equation of Line :
y = mx + c
Put the point (7/5 , 1/5) and slope m = -3/4 in equation to find the value of c
1/5 = (-3/4)(7/5) + c
c = 5/4
y = -3x/4 + 5/4

Question no. 39
The equation of two lines through ( 0,a) which are at distance 'a' from the point (2a,2a) are
looks_one y-a=0 and 4x-3y -3a=0
looks_two y-a=0 and 4x-3y+3a=0
looks_3 y-a=0 and 4x-3y +3a=0
looks_4 none of these
Option looks_3 y-a=0 and 4x-3y +3a=0
Solution :

Question no. 40
Equations of two straight lines passing through the point (3, 2) and making an angle 45° with the line x - 2y = 3 are
looks_one 3x+y +7=0 and x+3y+9=0
looks_two 3x-y-7=0 and x+3y-9=0
looks_3 x+3y-7=0 and x +3y +7=0
looks_4 none of these
Option looks_two 3x-y-7=0 and x+3y-9=0
Solution :
x - 2y = 3 ⇒ y = (1/2)x - (1/3)
Slope of above given line = 1/2


Case 1 : m = 3
y = 3x + c
(3,2) is the point
2 = 3*3 + c
c = -7
y = 3x - 7
Case 2 : m = -1/3
y = (-1/3)x + c
2 = (-1/3)*3 + c
2 = -1 + c
c = 3
y = -(1/3)x + 3
3y + x -9 = 0

Question no. 41
The equation to the straight line passing through the point (acos3θ , asin3θ) and perpendicular to the line xsecθ + ycosecθ = a , is
looks_one xcosθ - ysinθ = acos2θ
looks_two xcosθ + ysinθ = acos2θ
looks_3 xsinθ + ycosθ = acos2θ
looks_4 none of these
Option looks_one xcosθ - ysinθ = acos2θ
Solution :
y = mx + c
ycosecθ = -xsecθ + a
y = -x(tanθ) + asinθ
m1.m2 = -1
m1.(-tanθ) = -1
m1 = cotθ
y =mx + c
asin3θ = cotθ(acos3θ) + c
c = asin3θ - cotθ(acos3θ)
c = a(sin4θ - cos4θ)/sinθ
c = a(sin2θ -cos2θ)(sin2θ + cos2θ)/sinθ
c = -acos2θ/sinθ
y = xcotθ - acos2θ/sinθ

Question no. 42
Equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the x- axis, is
looks_one x√3 + y +8=0
looks_two x√3 - y =8
looks_3 x√3 + y =8
looks_4 x- √3y + 8=0
Option looks_4 x- √3y + 8=0
Solution :
p = 4
θ = 120o
x cosθ + y sinθ = p
x cos120o + y sin120o = 4
x(-1/2) + y(√3/2) = 4
x- √3y + 8=0

Question no. 43
The equation of the line passes through (a,b) and parallel to the line , is
looks_one
looks_two
looks_3
looks_4
Option looks_two
Solution :
Equation of line parallel to
it means , equation of line :

Question no. 44
The equation of line passing through the point of intersection of the lines 4x- 3y -1=0 and 5x - 2y -3 =0 and parallel to the line 2y -3x +2 =0, is
looks_one x-3y=1
looks_two 3x-2y=1
looks_3 2x-3y=1
looks_4 2x-y=1
Option looks_two 3x-2y=1
Solution :
4x- 3y -1=0 ......[1]
5x - 2y -3 =0 ......[2]
Solve [1] and [2]
x = 1 , y = 1
Equation of line :
2y -3x +c =0
(1,1) satisfy the equation
2(1) - 3(1) + c = 0
c = 1
equation of line : 2y -3x + 1 = 0

Question no. 45
The equation of line passing through point of intersection of lines 3x-2y-1= 0 and x-4y +3= 0 and the point (π, 0) , is
looks_one x-y =π
looks_two x-y = π(y+1)
looks_3 x-y = π(1-y)
looks_4 x+ y = π(1-y)
Option looks_3 x-y = π(1-y)
Solution :
3x-2y-1= 0 ......[1]
x-4y +3= 0 ......[2]
Solve [1] and [2]
x = 1 , y = 1
Equation of line passing through (1,1) and (π, 0)
y -0 = (1-0)(x-π)/(1-π)
y(1-π) = x - π
x-y = π(1-y)
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