Practice question On Straight Lines For NDA , Airforce and IITJEE 16-30

Straight Lines

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Question no. 16

The equations of the lines through the origin making an angle of 60° with the line x+√3y+3√3 =0 are

looks_one y=0, x - √3y=0

looks_two y=0, x + √3y=0

looks_3 x=0, x - √3y=0

looks_4 x=0, x + √3y=0

Option looks_3 x=0, x - √3y=0

Solution :
x+√3y+3√3 =0
√3y = -x - 3√3
m¹ = -1/√3

Case 1 :
√3 - m₂ = m₂ + 1/√3
m₂ = 1/√3
Equation of Line : y = mx [Lines passes through origin ]
y = x/√3
Case 2 : -√3 + m₂ = m₂ + 1/√3
m₂ will be infinity to satisfy the equation
therefore x = 0

Question no. 17

A straight line through P(1, 2) is such that its intercept between the axes is bisected at P. Its equation is

looks_one x + 2y=5

looks_two x - y + 1= 0

looks_3 x + y -3= 0

looks_4 2x + y -4 = 0

Option looks_4 2x + y -4 = 0

Solution :

Mid point
(0 + a)/2 = 1 ⇒ a = 2
(0 + b)/2 = 2 ⇒ b = 4
x/2 + y/4 = 1
2x + y - 4 = 0

Question no. 18

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

looks_one 1/3

looks_two 2/3

looks_3 1

looks_4 4/3

Option looks_4 4/3

Solution :
3x + y = 3 ⇒ y = -3x + 3
m₁ = -3
m₁.m₂ = -1
m₂ = 1/3
y = (1/3)x + c
(2, 2) is point on given line
2 = 2/3 + c ⇒ c= 4/3

Question no. 19

Equation of line passing through (1, 2) and perpendicular to 3x + 4y +5 =0 is

looks_one 3y=4x -2

looks_two 3y = 4x + 3

looks_3 3y = 4x +4

looks_4 3y = 4x + 2

Option looks_4 3y = 4x + 2

Solution :
3x + 4y + 5 = 0
y = (-3/4)x - 5/4
m₁ = -3/4
m₁.m₂ = -1
(-3/4)m₂ = -1 ⇒ m₂ = 4/3
y = (4/3)x + c
(1, 2) is point on given line
2 = 4/3 + c ⇒ c= 2/3
y = (4/3)x + 2/3
3y = 4x + 2

Question no. 20

The equation to the straight line passing through the point of intersection of the lines 5x - 6y-1=0 and 3x +2y + 5 = 0 and perpendicular to the line 3x - 5y +11=0 is

looks_one 5x +3y +8=0

looks_two 5x +3y + 11=0

looks_3 3x - 5y +8 =0

looks_4 3x - 5y +11=0

Option looks_one 5x +3y +8=0

Solution :
5x - 6y-1=0 ........[1]
3x +2y + 5 = 0 .......[2]
Solve [1] and [2]
x = -1 , y= -1
3x - 5y +11=0 ⇒ y = (3/5)x + 11/5
m₁ = 3/5
m₁.m₂ = -1
(3/5)m₂ = -1 ⇒ m₂ = -5/3
y = (-5/3)x + c
(-1, -1) is point on given line
-1 = 5/3 + c ⇒ c= -8/3
y = (-5/3)x - 9/3
3y = -5x -8

Question no. 21

In what ratio does the y axis divide the join of (-3,-4) and (1,-2)

looks_one 1:3

looks_two 2:3

looks_3 3:1

looks_4 none of these

Option looks_3 3:1

Solution :

(m - 3n)/(m+n) = 0
m -3n = 0
m : n = 3 : 1

Question no. 22

Three vertices of a parallelogram taken in order are ( -1, -6), (2,-5) and (7,2). The fourth vertex is

looks_one (1,4)

looks_two (4,1)

looks_3 (1,1)

looks_4 (4,4)

Option looks_two (4,1)

Solution :
Parallelogram diagonals bisect each other

(x+2)/2 = 3
x = 4
(y-5)/2 = -2
y = 1

Question no. 23

If the vertices of a triangle are A(1, 4) B(3,0) and C (2, 1) , then the length of the median passing through C is

looks_one 1

looks_two 2

looks_3 √2

looks_4 √3

Option looks_one 1

Solution :
Median divides the base into two equal parts

length of the median passing through C
= [(2-2)² + (2-1)²]^1/2
= 1

Question no. 24

The coordinates of the join of trisection of the points (-2,3),(3,-1) nearer to (-2,3 ) is

looks_one (-1/3, 5/3)

looks_two (4/3 , 1/3)

looks_3 (-3/4 ,2)

looks_4 (1/3, 5/3)

Option looks_one (-1/3, 5/3)

Solution :

x = (1*3 + 2 *(-2) ) / (2 + 1) = -1/3
y = (1*(-1) + 2 *(3) ) / (2 + 1) = 5/3

Question no. 25

The ratio in which x-axis divides the join of the points (2, -3) and (5,6) is

looks_one 2:1

looks_two 1:2

looks_3 2:-1

looks_4 none of these

Option looks_two 1:2

Solution :

Question no. 26

If the middle point of the line segment joining the points (5,a) and (b,7) be (3,5) , then (a,b) =

looks_one (3,1)

looks_two (1,3)

looks_3 (-2,-2)

looks_4 (-3,-1)

Option looks_one (3,1)

Solution :

3 = (5+b)/2 ⇒ b = 1
5 = (a+7)/2 ⇒ a = 3

Question no. 27

If the point dividing internally the line segment joining the points (a,b) and (5,7) in the ratio 2:1 be (4,6) , then

looks_one a=1 , b=2

looks_two a=2 , b=-4

looks_3 a=2 , b=4

looks_4 a=-2 , b=4

Option looks_3 a=2 , b=4

Solution :

(10 + a)/3 = 4 ⇒ a = 2
(14+b)/3 = 6 ⇒ b = 4

Question no. 28

The distance of the point (bcosθ , bsinθ) from origin is

looks_one bcotθ

looks_two b

looks_3 btanθ

looks_4 b√2

Option looks_two b

Solution :
A(bcosθ , bsinθ) is a point
Distance between Origin and A
OA = [(bcosθ - 0)² + (bsinθ - 0 )²]^1/2
OA = [b²cos²θ + b²sin²θ ]^1/2
OA = [b²]^1/2 = b

Question no. 29

The length of altitude through A of the triangle ABC where A(-3,0) B(4,-1) C (5,2) is

looks_one 2/√10

looks_two 4/√10

looks_3 11/√10

looks_4 22/√10

Option looks_4 22/√10

Solution :

BC = √10
Area of triangle = (1/2)[x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ] = 11
Area = (1/2)*BC*AD
11 = (1/2)*√10*AD
AD = 22/√10

Question no. 30

If the coordinates of vertices of ∆OAB are (0,0) ,(cosα, sinα) and (-sinα ,cosα) respectively , then OA² + OB² =

looks_one 0

looks_two 1

looks_3 2

looks_4 3

Option looks_3 2

Solution :
OA² = (0- cosα)² + (0 - sinα)²
OA² = cosα² + sinα² ..........[1]
OB² = (0 + sinα)² + (0 - cosα)²
OA² = sinα² + cosα² ..........[2]
Adding [1] and [2]
OA² + OB² = cosα² + sinα² + sinα² + cosα²
OA² + OB² = 2

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