Straight Lines Test -7

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Straight Lines Test -7

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Weekly Quiz Competition

Question 1
1 / -0

The area of the quadrilateral formed by the lines | x | + | y | = 1 is

A
2

B
4

C
8

D
none of these

Solution

Equations of the lines are

x≥0, y≥0, then x+y=1

x≤0, y≤0, then x+y=-1

x≥0, y≤0, then x-y = 2

x≤0 , y≥0, then x-y=-2

Clearly these lines form a square, whose coordinates are (1,1),(1,-1),(-1,-1),(-1,1)

Hence its area is 4 x [1/2]1 x 1= 2 sq units
Question 2
1 / -0

If a , b , c are in A . P. then straight line ax + by + c = 0 will always pass through a fixed point whose coordinates are

A
( 1 , - 2 )

B
( - 1 , - 2 )

C
( 1 , 2 )

D
none of these

Solution

Since a,b,c are in A.P,

a+c = 2b

This implies a-2b+c = 0

This implies the the family of lines is concurrent at (1,2)
Question 3
1 / -0

If the area of the triangle formed by $$ (-2,5), (x,-3) $$ and $$(3,2)$$ is $$14 $$ square units, then $$x=$$ ____.

A
1

B
2

C
-2

D
-1

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
$$\therefore \displaystyle Area=\frac{1}{2}\left [ -2\left ( -3-2 \right )+x\left ( 2-5 \right )+3\left ( 5+3 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left [ 10-3x+24 \right ]=\frac{34-3x}{2}$$
$$\displaystyle \therefore \frac{34-3x}{2}=14$$ or x = 2
Question 4
1 / -0

The distance between the points (sin x, cos x) and (cos x -sin x) is

A
$$1$$

B
$$\displaystyle \sqrt{2}$$

C
$$2 sin x cos x$$

D
$$4 sin x cos x$$

E
none of these

Solution

Distance between $$(sinx,cosx)$$ and $$(cosx,-sinx)$$ is
$$r=\sqrt{(sinx-cox)^2+(cosx-(-sinx)^2)}$$
$$=>r=\sqrt{(sin^2x+cos^2x-2sinxcosx)+(cox^2x+sin^2x-2sinxcosx)}$$
$$=>r=\sqrt{1+1}$$
$$=>r=\sqrt{2}$$
Question 5
1 / -0

Find the distance from the point (2, 3) to the line 3x + 4y + 9 = 0

A
$$5$$

B
$$5.4$$

C
$$5.8$$

D
$$6.2$$

E
none of these

Solution

Distance from point $$(2,3)$$ to the line $$3x+4y+9=0$$

$$=>r=\dfrac{|3(2)+4(3)+9|}{\sqrt {3^2+4^2}}$$

$$=>r=\dfrac{|16+12+9|}{\sqrt {9+16}}$$

$$=>r=\dfrac{27}{5}$$

$$=>r=5.4$$
Question 6
1 / -0

If the area of the triangle formed by the points $$(-2,3), (4,-5)$$ and $$(-3,y)$$ is 10 square units then $$y =$$

A
$$\dfrac{19}{3}$$

B
$$\dfrac{21}{3}$$

C
$$\dfrac{23}{3}$$

D
$$\dfrac{25}{3}$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
$$\displaystyle \text{Area} =\frac{1}{2}\left [ -2\left ( -5-y \right )+4\left ( y-3 \right )-3\left ( 3+5 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left ( 10+2y+4y-12-24 \right )$$
$$\displaystyle =\frac{1}{2}\left ( 6y-26 \right )=3y-13$$
$$\displaystyle 3y-13=10\ \mathrm{sq. unit}$$
$$\displaystyle y=\frac{23}{3}$$
Question 7
1 / -0

What is the slope of the line 3x + 2y + 1 = 0?

A
3/2

B
2/3

C
-3/2

D
-2/3

Solution

The given equation of the line is
$$3x+2y+1=0$$
$$=>3x+2y=-1$$
$$=>2y=-3x-1$$
$$=>y=\frac{-3}{2}x-\frac{-1}{2}$$
Here $$m=\frac{-3}{2}$$
Thus, slope of the line is $$\frac{-3}{2}$$
Question 8
1 / -0

Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.

A
3 square units

B
4 square units

C
5 square units

D
6 square units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
The three vertices of the triangle are $$A(1,1), B(3,-3), C(5,-3)$$
Area of triangle $$=\dfrac{|1(-3-(-3)+3(-3-1)+5(1-(-3))|}{2}$$
$$=\dfrac{|1(0)+3(-4)+4(4)|}{2}$$
$$=\dfrac{|-12+20|}{2}$$
$$=\dfrac{8}{2}$$
$$=4$$ square units.
Question 9
1 / -0

The area of triangle whose vertices are $$A (-3, -1), B(5, 3)$$ and $$C(2, -8)$$ is ____ $$\text{ sq. units}$$.

A
$$34\text{ sq. units}$$

B
$$36\text{ sq. units}$$

C
$$38\text{ sq. units}$$

D
$$32\text{ sq. units}$$

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

The given vertices of the triangle are $$A(-3,-1), B(5,3)$$ and $$C(2,-8)$$.
So, by using the above formula,
$$\begin{aligned}{}\text{Area of the triangle} &= \frac{1}{2} {[ - 3(3 - ( - 8)) + 5( - 8 - ( - 1)) + 2( - 1 - 3)]}\\ &= \frac{1}{2}{[ - 33 - 35 - 8]}\\ &= \frac{{[ - 76]}}{2}\\& = \frac{{76}}{2}\quad\quad\quad\quad\dots[\text{Area can never be negative so, we ignore negative sign}]\\& = 38\text{ sq. units}\end{aligned}$$

So, the area of the triangle is equal to $$38\text{ sq. units}$$.
Question 10
1 / -0

If the inclination of a line is $$45^\circ$$, then the slope of the line is ?

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$2$$

Solution

The inclination of line is $$45^0$$
Hence slope of line $$m=\tan 45^0=1$$
Question 11
1 / -0

The two diagonally opposite vertices of a square are $$(6,\,6)$$ and $$(0,\,0)$$. Find the point which lies on X-axis.

A
$$(6,0)$$

B
$$(0,6)$$

C
$$6,6$$

D
None

Solution

$$OA=AB=BC=OC$$ ....Sides of a square
By distance formula,
$$AB=\sqrt{(6-x)^2+(0-6)^2}$$
$$36=(6-x)^2+36$$
$$x=6$$
Hence, the point lying on the x-axis is $$A \equiv (6,0)$$.
Question 12
1 / -0

One line passes through the points $$(1,9)$$ and $$(2,6)$$ another line passes through $$(3, 3)$$ and $$(-1, 5)$$ The acute angle between the two lines is

A
30$$\displaystyle ^{\circ}$$

B
45$$\displaystyle ^{\circ}$$

C
60$$\displaystyle ^{\circ}$$

D
135$$\displaystyle ^{\circ}$$

Solution

Slope of line passing through $$(1,9)$$ and $$(2,6)$$ is
$$m_1=\dfrac{6-9}{2-1}$$
$$=\dfrac{-3}{1}$$
$$=-3$$
$$=>tan\theta_1=-3$$
$$=>\theta_1=tan^{-1}(-3)$$
$$=-71.57^0$$
Slope of line passing through $$(3,3)$$ and $$(-1,5)$$ is
$$m_2=\dfrac{5-3}{-1-3}$$
$$=\dfrac{-2}{4}$$
$$=\dfrac{-1}{2}$$
$$=>tan\theta_2=\dfrac{-1}{2}$$
$$=>\theta_2={ tan }^{ -1 }\dfrac { -1 }{ 2 } $$
$$\theta_2=-26.57^0$$
Angle between them $$=-26.57^0+71.57^0$$
$$=45^0$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test -7

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Straight Lines Test -7

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SHARING IS CARING

Question 1

2

4

8

none of these

Solution

Question 2

( 1 , - 2 )

( - 1 , - 2 )

( 1 , 2 )

none of these

Solution

Question 3

1

2

-2

-1

Solution

Question 4

$$1$$

$$\displaystyle \sqrt{2}$$

$$2 sin x cos x$$

$$4 sin x cos x$$

none of these

Solution

Question 5

$$5$$

$$5.4$$

$$5.8$$

$$6.2$$

none of these

Solution

Question 6

$$\dfrac{19}{3}$$

$$\dfrac{21}{3}$$

$$\dfrac{23}{3}$$

$$\dfrac{25}{3}$$

Solution

Question 7

3/2

2/3

-3/2

-2/3

Solution

Question 8

3 square units

4 square units

5 square units

6 square units

Solution

Question 9

$$34\text{ sq. units}$$

$$36\text{ sq. units}$$

$$38\text{ sq. units}$$

$$32\text{ sq. units}$$

Solution

Question 10

$$0$$

$$-1$$

$$1$$

$$2$$

Solution

Question 11

$$(6,0)$$

$$(0,6)$$

$$6,6$$

None

Solution

Question 12

30$$\displaystyle ^{\circ}$$

45$$\displaystyle ^{\circ}$$

60$$\displaystyle ^{\circ}$$