Straight Lines Test -3

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Straight Lines Test -3

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the line which passes through the point ( 1 , - 2 ) and cuts off equal intercepts from the axis is

A
x – y – 2 = 0

B
x – y - 1 = 0

C
x + y + 1 = 0

D
x + y = 1

Solution

Let the equation of the line which has equal intercepts be

x/a + y/a =1

Substituting the values for x and y

1 -2 =a

Therefore a = -1

Hence the equation of the line is x+y+1=0

Option 1 is correct
Question 2
1 / -0

If a line is drawn through the origin and parallel to the line x – 2y + 5 = 0 , its equation is

A
x = 2y

B
2x + y = 0

C
x – 2y – 5 = 0

D
none of these

Solution

If the line is parallel to the given line x-2y + 5 =0,

then the required line will have same slope

Hence the equaion of the given line is x-2y+k=0

since it passes through the origin,

0+0+k=0

Therefore k=0

Hence the equation of the required line is x-2y=0 or x=2y

Option 1 is correct
Question 3
1 / -0

The line passing through ( 1, 1 ) and parallel to the line 2x – 3y + 5 = 0 is

A
3x – 2y = 1

B
2x – 3y + 1= 0

C
2x + 3y = 5

D
3x + 2y = 5

Solution

The required line is parallel to the given line, hence it has same slope

Therefore the equation of the required line is 2x-3y+k=0

Since it passes through (1,1)

2(1) - 3(1) +k = 0

Therefore k=1

Hence the equation of the required line is 2x - 3y + 1=0

Hence option 1 is correct
Question 4
1 / -0

The equation of the line which passes through the point ( 2 , - 3 ) and cuts off equal intercepts from the axis is

A
x – y – 2 = 0

B
x – y = 1

C
x + y = 1

D
x + y + 1 = 0

Solution

The equation of the line which cuts equal intercepts is

x+y=a

Since it passes through (2,-3),

2 - 3 =a

Hence a = -1

Hence the equation of the required line is x + y +1 = 0

Hence option 1 is correct
Question 5
1 / -0

The straight lines x + y - 4 = 0 , 3x + y – 4 = 0 , x - 3y – 4 = 0 form a triangle which is

A
obtuse angled triangle

B
equilateral

C
right angled triangle

D
none of these

Solution

The triangle formed by these lines is a right angled triangle

If the lines are perpendicular to each other, then the product of their slopes is -1

The slope of lines 3x + y – 4 = 0 , x - 3y – 4 = 0 are -3 and 1/3 respectively.

The product of the slopes is -1

Hence these two lines are perpendicular to each other

This infers that the triangle formed by these lines is a right angled triangle.
Question 6
1 / -0

The line which passes through the point ( 0 , 1 ) and perpendicular to the line x – 2y + 11 = 0 is

A
2x + y – 1 = 0

B
2x – y + 3 = 0

C
2x – y + 1 = 0

D
none of these

Solution

The line which is perpendicular to the given line is 2x + y + k = 0

Since it passes through (0,1)

2(0) + 1 + k = 0

This implies k = -1

Hence the equation of the required line is 2x + y - 1 = 0
Question 7
1 / -0

The acute angle between the lines x – y = 0 and y = 0 is

A
60⁰

B
45⁰

C
75⁰

D
30⁰

Solution

Slope of the line x - y = 0 is 1. This imlpies the tanθ = 1 Hence the angle made by the line with X- axis is 45⁰

The angle made by the line y = 0 with the Y axis is 90⁰

Therefore the acute angle between the lines is 45⁰

Option 1 is correct.
Question 8
1 / -0

The middle point of the line segment joining $$ (3 , 1) $$ and $$ (1 , 1) $$ is shifted by two units ( in the sense increasing y) perpendicular to the line segment. Then the coordinates of the point in the new position is

A
$$(2,4)$$

B
$$(1,2)$$

C
$$(1,4)$$

D
$$(2,3)$$

Solution

Mid point of $$(3,1) (1,1) $$ is $$(2,1)$$
on increasing 2 units in terms of y it becomes $$(2,3)$$
Question 9
1 / -0

A line passes through the point $$(3,4)$$ and cut off intercepts from the coordinates axes such that their sum is $$14$$. The equation of the line is

A
$$4x-3y=24$$

B
$$4x+3y=24$$

C
$$3x-4y=24$$

D
$$3x+4y=24$$

Solution

Let the intercept by the line on the axes are $$a$$ and $$(14-a)$$
Thus the equation,
$$\dfrac{x}{a}+\dfrac{y}{14-a}=1$$ and passes through $$(3,4)$$ ----- (1)

$$\dfrac{3}{a}+\dfrac{4}{(14-a)}=1$$

Therefore,
$$a=6,b=8$$
So the equation is
Put in (1)
$$4x+3y=24$$
Question 10
1 / -0

The points $$A(-4,1)$$, $$B(-2,-2),C(4,0),D(2,3)$$ are the vertices of

A
parallelogram

B
rectangle

C
rhombus

D
None of these

Solution

$$AB=\sqrt{(-4+2)^{2}+(1+2)^{2}}=\sqrt{13}$$
$$BC=\sqrt{(-2-4)^{2}+(-2-0)^{2}}=2\sqrt{10}$$
$$CD=\sqrt{(4-2)^{2}+(0-3)^{2}}=\sqrt{13}$$
$$DA=\sqrt{(-4-2)^{2}+(1-3)^{2}}=2\sqrt{10}$$
Opposite sides are equal
$$\implies $$ it must be a parallelogram
Question 11
1 / -0

The points $$A(2a,4a),B(2a,6a)$$ and $$C(2a+\sqrt{3}a,5a)$$ (when $$a> 0$$) are vertices of

A
an obtuse angled triangle

B
an equilateral triangle

C
an isosecels obtuse angled triangle

D
a right anged triangle

Solution
Question 12
1 / -0

Which of the following sets of points form an equilateral triangle

A
$$\left( {1,0} \right),\left( {4,0} \right),\left( {7, - 1} \right)$$

B
$$\left( {\dfrac{2}{3},0} \right),\left( {0,\dfrac{2}{3}} \right),\left( {1,1} \right)$$

C
$$\left( {0,0} \right),\left( {\dfrac{3}{2},\dfrac{4}{3}} \right),\left( {\dfrac{4}{3},\dfrac{3}{2}} \right)$$

D
$$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {0,2\sqrt 3 } \right)$$

Solution
Question 13
1 / -0

A line is of length $$10$$ m and one end is $$(2,-3)$$, the $$x$$ - co-ordinate of the other is $$8$$, then its $$y$$- coordinate is:

A
$$5,-11$$

B
$$-5,11$$

C
$$3,11$$

D
$$2,3$$

Solution

Let the $$y$$-coordinate of the other end be $$\alpha$$
Therefore,
coordinat of other end $$= \left( 8, \alpha \right)$$
Now, distance between $$\left( 2, -3 \right)$$ and $$\left( 8, y \right)$$-
$$d = \sqrt{{\left( 8 - 2 \right)}^{2} + {\left( y - \left( -3 \right) \right)}^{2}}$$
Given that length of line is $$10$$.
$$\therefore \sqrt{{\left( 8 - 2 \right)}^{2} + {\left( y - \left( -3 \right) \right)}^{2}} = 10$$
Squaring both sides we have
$$36 + \left( {y}^{2} + 9 + 6 y \right) = 100$$
$$36 + {y}^{2} + 18y + 9 = 100$$
$$\Rightarrow {y}^{2} + 6y - 55 = 0$$
$$\Rightarrow {y}^{2} + 11y - 5y - 55 = 0$$
$$\Rightarrow \left( y + 11 \right) \left( y - 5 \right) = 0$$
$$\Rightarrow y = 5, -11$$
Hence the $$y$$-coordinate will be $$5, -11$$.
Hence the correct answer is $$\left( A \right) 5, -11$$.
Question 14
1 / -0

The coordinates of a point on the line y=x where perpendicular from the line 3x+4y=12 is 4 units, are

A
$$(\dfrac{3}{7},\dfrac{5}{7})$$

B
$$(\dfrac{3}{2},\dfrac{3}{2})$$

C
$$(-\dfrac{8}{7},-\dfrac{8}{7})$$

D
$$(\dfrac{32}{7},-\dfrac{32}{7})$$

Solution

Let any point on line $$y\, = x$$
from point of P(a,a),
Distance from point,
$$\begin{array}{l} \left| { \dfrac { { 3a+4a-12 } }{ { \sqrt { 9+16 } } } } \right| =4 \\ \left| { \dfrac { { 7a-12 } }{ 5 } } \right| \, \, =4,\, \, \, \dfrac { { 7a-12 } }{ 5 } =\pm 4 \\ we\, have\, two\, conditions, \\ 7a-12=20-----(1) \\ a=\dfrac { { 32 } }{ 7 } \\ 7a-12=-20------(2) \\ a=\dfrac { { -8 } }{ 7 } \\ Coordinate\, of\, po{ { int } }\, is\, a, \\ then, \\ P\left( { \dfrac { { 32 } }{ 7 } ,\dfrac { { 32 } }{ 7 } } \right) \, \, and\, \, P\left( { \dfrac { { -8 } }{ 7 } ,\dfrac { { -8 } }{ 7 } } \right) \\ So,the\, \, correct\, option\, is\, C. \end{array}$$
Question 15
1 / -0

The mid points of three sides of a triangle are (0, 1) (0, 2) and (0, 3). Area of this triangle.

A
3

B
0

C
1

D
None of these

Solution
Question 16
1 / -0

lf the line joining the points $$(\mathrm{a}\mathrm{t}_{1}^{2},2\mathrm{a}\mathrm{t}_{1}),(\mathrm{a}\mathrm{t}_{2}^{2},\ 2\mathrm{a}\mathrm{t}_{2})$$ is parallel to $$\mathrm{y}=\mathrm{x},$$ then $$\mathrm{t}_{1}+\mathrm{t}_{2}=$$

A
$$\displaystyle \frac{1}{2}$$

B
$$4$$

C
$$\displaystyle \frac{1}{4}$$

D
$$2$$

Solution

Slope of line joining the points $$(at_{1}^2, 2at_{1}), (at^2_2, 2 at_2)$$ is $$m=\dfrac{2a(t_2-t_1)}{a(t_2^2-t_1^2)}$$
$$\Rightarrow m=\dfrac{2}{(t_1+t_2)}$$

Now, given line is $$y=x$$
Slope of given line is $$1.$$
Since,the line joining given points is parallel to $$y=x$$
So, their slopes are equal
$$ \dfrac{2}{t_1+t_2}=1$$
$$t_1+t_2=2$$
Question 17
1 / -0

The acute angle between the lines
$$l x + my =l+m, l(x-y) +m(x + y)=2m $$ is

A
$$\displaystyle \frac{\pi}{4}$$

B
$$\displaystyle \frac{\pi}{6}$$

C
$$\displaystyle \frac{\pi}{2}$$

D
$$\displaystyle \frac{\pi}{3}$$

Solution

$$lx+my=l+m$$ & $$l(x-y)+m(x+y)=2m$$
$$ \displaystyle \Rightarrow m_1 = -\frac{l}{m},\; m_2=\frac{m+l}{l-m}$$
$$ \displaystyle \tan\theta=\left | \frac{m_1-m_2}{1+m_2m_2} \right| $$
$$ \displaystyle =\left| \frac { \left( -\frac { l }{ m } -\frac { m+l }{ l-m } \right) }{ 1+\left( -\frac { l }{ m } \right) \left( \frac { m+l }{ l-m } \right) } \right| \\ \displaystyle =\left| \frac { -{ l }^{ 2 }+ml-{ m }^{ 2 }-ml }{ ml-{ m }^{ 2 }-ml-{ l }^{ 2 } } \right| $$
$$ \displaystyle \tan\theta=1$$
$$ \displaystyle \Rightarrow \theta=\frac{\pi}{4}$$

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Straight Lines Test -3

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Straight Lines Test -3

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SHARING IS CARING

Question 1

x – y – 2 = 0

x – y - 1 = 0

x + y + 1 = 0

x + y = 1

Solution

Question 2

x = 2y

2x + y = 0

x – 2y – 5 = 0

none of these

Solution

Question 3

3x – 2y = 1

2x – 3y + 1= 0

2x + 3y = 5

3x + 2y = 5

Solution

Question 4

x – y – 2 = 0

x – y = 1

x + y = 1

x + y + 1 = 0

Solution

Question 5

obtuse angled triangle

equilateral

right angled triangle

none of these

Solution

Question 6

2x + y – 1 = 0

2x – y + 3 = 0

2x – y + 1 = 0

none of these

Solution

Question 7

600

450

750

300

Solution

Question 8

$$(2,4)$$

$$(1,2)$$

$$(1,4)$$

$$(2,3)$$

Solution

Question 9

$$4x-3y=24$$

$$4x+3y=24$$

$$3x-4y=24$$

$$3x+4y=24$$

Solution

Question 10

parallelogram

rectangle

rhombus

None of these

Solution

Question 11

an obtuse angled triangle

an equilateral triangle

an isosecels obtuse angled triangle

a right anged triangle

Solution

Question 12

$$\left( {1,0} \right),\left( {4,0} \right),\left( {7, - 1} \right)$$

$$\left( {\dfrac{2}{3},0} \right),\left( {0,\dfrac{2}{3}} \right),\left( {1,1} \right)$$

$$\left( {0,0} \right),\left( {\dfrac{3}{2},\dfrac{4}{3}} \right),\left( {\dfrac{4}{3},\dfrac{3}{2}} \right)$$

$$\left( {0,0} \right),\left( {3,\sqrt 3 } \right),\left( {0,2\sqrt 3 } \right)$$

Solution

60⁰

45⁰

75⁰

30⁰