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Competition Level Test: Straight Line- 3


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30 Questions MCQ Test Mathematics For JEE | Competition Level Test: Straight Line- 3

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Competition Level Test: Straight Line- 3 - Question 1

Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 1

AT FIRST FIND THE COMMON PONT OF THIS FAMILY WHICH IS A (-1/3 , 1/3)
THEN FOR THE LINE TO BE AT THE FARTHEST DISTANCE FROM B(1,-3) THE LINE SHOULD BE PREPENDICULAR TO AB THEREFORE THE SLOPE OF LINE m= 2/5
EQUATION OF LINE IS 2x + 2/3 = 5y - 5/3 WHICH IS HENCE 6x-15y-7=0

Competition Level Test: Straight Line- 3 - Question 2

One side of an equilateral triangle is 3x+4y=7 and its vertex is (1,2). Then the length of the side of the triangle is

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 2

Competition Level Test: Straight Line- 3 - Question 3

Find the equation of the bisector of the angle between the lines x+2y–11 = 0, 3x–6y–5 = 0 which contains the point (1,–3)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 3

x+2y−11=0    ..(1)

3x−6y−5=0    ..(2)

Point of intersection of (1) and (2)

x=19/3

y=7/3

Equation passing through (19/3,7/3) and (1,-3)

y-7/3=1616(x-19/3)

x-y-4=0

Competition Level Test: Straight Line- 3 - Question 4

A straight line which make equal intercepts on +ve x and y axes and which is at a distance '1' unit from the origin intersects the straight line y = 2x + 3 + √2 at (x0, y0) then 2x0 + y0

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 4

Equation of the straight line having equal intercepts is x + y = k and proceed.

Competition Level Test: Straight Line- 3 - Question 5

The locus of the image of the point (2, 3) in the line (x -2y +3) + λ (2x -3y + 4) = 0 (λ∈R) is 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 5

A(2,3),B(1,2),P(x,y)
AB = BP

Competition Level Test: Straight Line- 3 - Question 6

A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q. As L varies, the absolute minimum value of OP+OQ is (O is origin)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 6

The equation of the line L be x coordinates of P and Q are 3x + 4y = 9 and y = mx + 1 and P (-3, 4) .
So Q (0, 1)
So, absolute minimum value of x

Competition Level Test: Straight Line- 3 - Question 7

The number of integral values of m for which the x coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is  

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 7

Solving given equations we get  is an integer, if
(-2, 0) y = mx So m has two integral values.

Competition Level Test: Straight Line- 3 - Question 8

Area of the parallelogram formed by the lines y = mx, y = mx+ 1, y = nx and y = nx+ 1 equals (m ≠ n)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 8

If  be the distance between parallel sides and θ be the angle between adjacent sides, then

Requried area  Whre  y = mx + 1, x (distance between | | lines) and π/4 (0,3) 

Competition Level Test: Straight Line- 3 - Question 9

If the midpoint of sides of a triangle are (a, 0), (0, b) and (-a, -b), then the centroid of the triangle is

Competition Level Test: Straight Line- 3 - Question 10

The triangle ABC has medians AD, BE, CF. AD lies along the line y = x + 3, BE lies along the line y = 2x + 4, AB has length 60 and angle C = 90°, then the area of ΔABC is

Competition Level Test: Straight Line- 3 - Question 11

Let A(1, 0), B(0, –2) and P(a, b) be a third point that lies on the straight line 'x - y= 0' such that, |PA + PB| is minimum then the values of |a + b| =

Competition Level Test: Straight Line- 3 - Question 12

Let A(0, 0), B(3, 0), C(2, 1) then area of the square inscribed in the triangle such that, one side of the square is taken along 

Competition Level Test: Straight Line- 3 - Question 13

Let O(0, 0), A(–1, 1) the point ‘A’ is rotated 60o in anticlockwise sense about origin then the slope of line joining original and the new positions of ‘A’ is 

Competition Level Test: Straight Line- 3 - Question 14

Let y = 2x + c1, y = 2x + c2 meet X-axis at A1, A2 respectively and Y-axis at B1, B2 respectively then the locus of the point of intersection of lines 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 14


Competition Level Test: Straight Line- 3 - Question 15

Let A(2, 3) and B is the image of A with respect to the lines (2x - 3y + 4) + k (x - 2y + 3) = 0   then the locus of B is 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 15

The given line is 
(2x - 3y + 4) + k (x - 2y + 3) = 0  
as x−2y=−3
2x−3y=−4
multiplying eqn(2) by 2 and subtracting we get, y=2
so, x=−3+4=1
so point of intersection P is (1,2)
A′ is the reflection of A
as, AP=BP=A′P
so, √(2-1)2+(3-2)2] =[(h-1)2+(k-2)2]1/2
1+1=h2+k2-2h-4k+1+4
h2+k2−2h−4k+3=0
x2+y2-2x-4y+3=0

Competition Level Test: Straight Line- 3 - Question 16

Let 'P' be a point inside the equilateral triangle ABC, PA = 3, PB = 4, PC = 5 then the area of the triangle ABC (in square units)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 16


Rotate the triangle by 60o


AC2 = AP2 + PC2 - 2AP.PCcos(θ + 60°)

Competition Level Test: Straight Line- 3 - Question 17

Let a straight line passing through P(1, 4) with negative slope cuts the coordinate axes at A, B then the area of the triangle OAB when OA + OB is minimum is

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 17

OA + OB = 1 + 4cot θ + 4 + tan θ
= 5 + tan θ + 4 cot θ
Attains minima only if tan θ = 4 cot θ

Competition Level Test: Straight Line- 3 - Question 18

The image of the curve  with respect to the straight line x + y= 1 is 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 18

Replace x by 1 – Y
y by 1 – X

Competition Level Test: Straight Line- 3 - Question 19

Let (a, b) lies on obtuse angle bisector of x + 2y + 1 = 0, 2x + y + 1 = 0 then the least value of (a - 1)2 + (b - 2)2 is 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 19

Find short dist. From (1, 2)

Competition Level Test: Straight Line- 3 - Question 20

Let A(0, 0) and x + y - 1 = 0, 2x + 1 = 0 are the altitude, median from the vertices B, C respectively of ΔABC then the area of the triangle 

Competition Level Test: Straight Line- 3 - Question 21

Let (x - 1)2 + (y - 2)2 = 1/4 then the minimum value of " x + 3y+ 4"

Competition Level Test: Straight Line- 3 - Question 22

Two vertical poles 20 mts, and 80 mts stand apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of others (in metres)

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 22

 The height "a" of the point E of intersection of the lines AD & BC drawn from the tops B & D of two poles AB & CD , to the feet of the other pole is given by LADDER THEOREM .
1/a = 1/AB + 1/CD
1/a = 1/20 + 1/80
= 5/80 = 1/16
So. a = 16 m

Competition Level Test: Straight Line- 3 - Question 23

A(3, 1), B(13, 6), C(13, 21), D(3, 16) forms a parallelogram, then the y-intercept of the line passing through (1, 0) and dividing the parallelogram into two equal areas 

Competition Level Test: Straight Line- 3 - Question 24

If the straight lines ax + by + p = 0 and x cosa + y sina = p are inclined at an angle p/4 and concurrent with the straight line x sina – y cosa = 0, then the value of a2 +b2 is

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 24

ax+by+p=0   ...(1)
⇒by=−ax+p
⇒y=−a/bx+p/b
So, Slope,m1 = −a/b 
and xcosθ+ysinθ−p=0    ..(2)
⇒ysinθ=−xcosθ+p
⇒y=−cosθ/sinθx+p/sinθ
So, Slope,m2=−cosθ/sinθ
Now, straight lines (1) and (2) include an angle π/4
i,e tanπ/4 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒ 1 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒1+m1m2|=|m1−m2|
⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣
=∣∣−a/b−(−cosθ/sinθ)∣∣
⇒∣∣(bsinθ+acosθ)/bsinθ∣∣
=∣∣(−asinθ+bcosθ)/bsinθ∣∣
⇒|bsinθ+acosθ|=|−asinθ+bcosθ|
Squaring both sides, we get
b2sin2θ+a2cos2θ+2absinθ cosθ=a2sin2θ+b2cos2θ−2absinθ cosθ    ...(3)
Given three lines are concurrent
Now, we will find the point of intersection of 
xcosθ+ysinθ=p and xsinθ−ycosθ=0
which is x=pcosθ and y=psinθNow, point of intersection will satisfy (1)
i.e apcosθ+bpsinθ+p=0 
⇒apcosθ+bpsinθ=−p
⇒acosθ+bsinθ=−1
Squaring both sides, we get
⇒a2cos2θ+b2sin2θ+2absinθ cosθ=1   ...(4)
Put the value of (4) in (3), we get
a2sin2θ+b2cos2θ−2absinθ cosθ =1   ....(5)
Adding (4) and (5), we get
a2cos2θ+b2sin2θ+2absinθ cosθ+a2sin2θ+b2cos2θ−2absinθ cosθ=1+1
⇒a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=2
⇒a2+b2=2

Competition Level Test: Straight Line- 3 - Question 25

Let (1, 2), (3, 8), (4, 1) are three vertices of a parallelogram the sum of all possible x coordinates of the fourth vertex of the parallelogram is

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 25

Let ABCD be a parallelogram.
then A (1,2), B (3,8) and C (4,1). Let the coordinates of D be (x,y)
Since the diagonals of a parallelogram bisect each other
Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
(1+4)/2 , (2+1)/2 = (3+x)/2 , (8+y)/2
5/2 , 3/2     (3+x)/2 , (8+y)/2
2.5, 1.5       , (3+x)/2 , (8+y)/2
(3+x)/2 = 2.5 , 3/2 = (8+y)/2
(3+x) = 5 ,     -5 = y
 2 = x        , -5 = y
∴ Coordinates of the fourth vertex are (2, -5)

Competition Level Test: Straight Line- 3 - Question 26

The orthocenter of the triangle formed by the lines x - 7y + 6 = 0, 2x - 5y - 6 = 0 and 7x + y - 8 = 0 is 

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 26

x-7y+6 = 0
m1 = -1/-7   = 1/7
7x+y-8 = 0 
m2 = -7/1
Slope = 1/7 * (-7)
            = 1
B is orthocentre of triangle ABC
x-7y = -6    ---------------------------(1)
7x+y = 8   ---------------------------(2)
y =8-7x    ----------------------------(3)
Put eq (3) in eq(1)
x-7(8-7x) = -6
x-56+49x = -6
50x = 50
x = 1
Similarly y = 1
Points are (1,1)

Competition Level Test: Straight Line- 3 - Question 27

The equations x = t3 + 9 and   represents a straight line where t is a parameter. Then y-intercept of the line is :

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 27

x=0
Put into x= t3 + 9
0 = t3 + 9
t3 = -9
y = 3(t3)/4 + 6
y = 3(-9)/4 + 6
y= -¾

Competition Level Test: Straight Line- 3 - Question 28

The vertices of triangle ABC are A(–1,–7), B(5, 1) and C(1, 4). The equation of the bisector of the angle ABC of ΔABC is :

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 28


Competition Level Test: Straight Line- 3 - Question 29

y - 1 = m1 x - 3 and y - 3 = m2 (x - 1) are two family straight lines (m1, m2R), at right angles to each other. The locus of their point of intersection is :

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 29


Locus of P (h, k) is circle with AB as diameter 
(x -3)(x -1) + (y -1)(y -3) = 0

Competition Level Test: Straight Line- 3 - Question 30

The equation of the line passing through the intersection of the lines 3x + 4y = -5, 4x + 6y = 6 and perpendicular to 7x - 5y + 3 = 0 is :

Detailed Solution for Competition Level Test: Straight Line- 3 - Question 30


Equation of altitude on BC
x + 4y= 13
Equation of altitude on AB
7x - 7y + 19 = 0
⇒ 

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