Straight Lines Test 21

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Straight Lines Test 21

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Weekly Quiz Competition

Question 1
1 / -0

If the slop of one of the lines represented by $$ax^2-6xy+y^2=0$$ is the square of the other,then the value of a is

A
$$-27$$ or $$8$$

B
$$-3$$ or $$2$$

C
$$-64$$ or $$27$$

D
$$-4$$ or $$3$$

Solution

$$m+m^2=6$$

$$m^3=a$$

$$m^3+m^6+3.m^3.6=6^3$$

$$a^2+19a-216=0$$

$$a=-27$$ or $$8$$
Question 2
1 / -0

Let $$A=(a_1, a_2)$$ and $$B=(b_1, b_2)$$ be two points in the plane with integer coordinates. Which one of the following is not a possible value of the distance between A and B?

A
$$\sqrt{65}$$

B
$$\sqrt{74}$$

C
$$\sqrt{83}$$

D
$$\sqrt{97}$$

Solution

Distance between $$A$$ and $$B$$ is =$$\sqrt {(a_1-b_1)^2+(a_2-b_2)^2}$$
It's given that the coordinates are integer, therefore $$(a_1-b_1)^2$$ and $$(a_2-b_2)^2$$ are the squares of integer numbers.
In options, we have
$$\sqrt{64}=\sqrt{8^2+1^2}$$
$$\sqrt{74}=\sqrt{7^2+5^2}$$
$$\sqrt{97}=\sqrt{9^2+4^2}$$
$$\sqrt{83}$$ cannot be written in this form.
Therefore, option C cannot the distance between A and B.
Question 3
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If the straight line $$ax + by + p = 0$$ and $$x\cos \alpha + y \sin \alpha = p$$ enclosed an angle of $$\dfrac {\pi}{4}$$ and the line $$x\sin \alpha - y \cos \alpha = 0$$ meets them at the same point, then $$a^{2} + b^{2}$$ is

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$x\sin \alpha-y\cos\alpha =0$$

=> $$x=\dfrac{y\cos\alpha }{\sin\alpha}$$

$$x\cos\alpha+y\sin\alpha=p$$

Put $$x=\dfrac{y\cos\alpha }{\sin\alpha}$$, we get

=> $$\dfrac{y\cos^2\alpha}{\sin\alpha}+y\sin\alpha=p$$

=>$$y=p\sin\alpha$$

Similiarly, $$x=p\cos\alpha$$

$$ax+by+p=0$$

Put $$x=p\cos\alpha$$ and $$y=p\sin\alpha$$

=> $$ap\cos\alpha+bp\sin\alpha+p=0$$

=> $$a\cos\alpha+b\sin\alpha=-1$$ ---- Equation 1

Slope of $$ax+by+p=0$$, $$m_1=-\dfrac{a}{b}$$

Slope of $$x\cos\alpha+y\sin\alpha=p$$, $$m_2=-\dfrac{\cos\alpha}{\sin\alpha}$$

Also $$\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$

=> $$\tan\dfrac{\pi}{4}=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$

=> $$|m_1-m_2|=|1+m_1m_2|$$

=> $$\left|\dfrac{a}{b}-\dfrac{\cos\alpha }{\sin\alpha}\right|=\left|1+\dfrac{a\cos\alpha}{b\sin\alpha}\right|$$

=> $$|a\sin\alpha-b\cos\alpha|=|b\sin\alpha+a\cos\alpha|$$

=> $$|a\sin\alpha-b\cos\alpha|=1$$ ---- Equation 2

Squaring and adding equation 1 and equation 2, we get

$$(a^2\cos ^2\alpha +b^2\sin ^2\alpha +2ab\sin \alpha \cos \alpha)+(b^2\cos ^2\alpha +a^2\sin ^2\alpha -2ab\sin \alpha \cos \alpha)=1+1$$

=> $$a^2+b^2=2$$

Option $$(C)$$
Question 4
1 / -0

The distance of the point $$A(a, b, c)$$ from the x-axis is

A
$$a$$

B
$$\sqrt {b^{2} + c^{2}}$$

C
$$\sqrt {a^{2} + b^{2}}$$

D
$$a^{2} + b^{2}$$

Solution

$$\textbf{Step-1: Apply distance formula to get the required unknown.}$$

$$\text{The given point is}$$ $$(a,b,c)$$

$$\text{The distance of the point}$$ $$(a,b,c) $$ $$\text{from x-axis will be}$$ $$\text{the perpendicular distance from point}$$

$$(a,b,c)$$ $$\text{to x-axis whose co-ordinates will be}$$ $$(a,0,0)$$

$$\text{Now, we have to use the distance formula}$$

$$\text{Distance=}$$ $$\sqrt{(a-a)^{2}+(b-0)^{2}+(c-0)^{2}}=\sqrt{b^{2}+c^{2}}$$ $$\textbf{[Using distance formula]}$$

$$\textbf{Hence, the correct option is B}$$
Question 5
1 / -0

$$ABC$$ is an isosceles triangle. If the coordinates of the base are $$B(1,3)$$ and $$C(-2,7)$$. The vertex $$A$$ can be

A
$$(1,6)$$

B
$$(-1/2,5)$$

C
$$(-7,1/6)$$

D
$$(5/6,6)$$

Solution

ABC-Isosceles triangle.
Co-ordinates of base are $$B(1, 3)$$ and $$C(-2, 7)$$
vertex 'A' can be any point on perpendicular bisector of BC, but not 'D'.
Slope of BC$$=\dfrac{7-3}{-2-1}=-\dfrac{4}{3}$$
Slope of Line(L)$$=\dfrac{-1}{\left(-\dfrac{4}{3}\right)}=\dfrac{3}{4}$$
Mid-point of BC$$-$$D$$=\left(\dfrac{-1}{2}, 5\right)$$
$$L: 3x-4y=3\left(-\dfrac{1}{2}\right)-4(5)$$
$$3x-4y=-\dfrac{3}{2}-20$$
$$6x-8y=-3-40$$
$$6x-8y+43=0$$
As, only $$(5/6, 6)$$ satisfies 'L'.
So, 'A' can be $$\left(\dfrac{5}{6}, 6\right)$$.
Question 6
1 / -0

How many points $$(x, y)$$ with integral co-ordinates are there whose distance from $$(1, 2)$$ is two units?

A
One

B
Two

C
Three

D
Four

Solution

$$PQ = 2\Rightarrow PQ^{2} = 4$$
$$(x - 1)^{2} + (y - 2)^{2} = 4$$
possibilities for integral coordinates
$$\Rightarrow (X - 1)^{2} + (y -2)^{2} = 4 + 0$$ or
$$= 0 + 4$$
Check $$(1) (x - 1)^{2} = 4\ (y - 2)^{2} = 0$$
$$\Rightarrow x - 1 = 2, -2\ ,y - 2 = 0$$
$$x = 3, -1\ ,y = 2$$
$$(3, 2)$$ and $$(-1, 2)$$
check $$(2) (x - 1)^{2} = 0\ ,(y - 2)^{2} = 4$$
$$x = 1\ ,y - 2 = 2, -2$$
$$x = 1\ ,y = 4, y = 0$$
$$(1, 4)$$ and $$(1, 0)$$
Total $$4$$ integral coordinates are possible.
Question 7
1 / -0

The distance between the points $$\left( a\cos { { 48 }^{ o } } ,0 \right) $$ and $$\left( 0,a\cos { { 12 }^{ o } } \right) $$ is $$d$$ then $${ d }^{ 2 }{ a }^{ 2 }=$$

A
$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +9 \right) }{ 4 } $$

B
$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 4 } $$

C
$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 8 } $$

D
$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +1 \right) }{ 8 } $$

Solution

Given $$A=(a\cos 48^o, 0)$$ & $$B=(0, a \cos 12^o)$$
$$\Rightarrow d^2=(a\cos 48-0)^2+(0-a\cos 12)^2$$
$$=a^2[\cos^248+\cos^212]$$
$$\cos^2\theta =\dfrac{1+\cos 2\theta}{2}$$ $$\Rightarrow a^2\left[\left(\dfrac{1+\cos 96}{2}\right)+\left(\dfrac{1+\cos 24}{2}\right)\right]$$
$$\Rightarrow a^2\left[\dfrac{2+\cos 96+\cos 24}{2}\right]$$
$$\left[\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos \left(\dfrac{A-B}{2}\right)\right]$$ $$\Rightarrow a^2\left[\dfrac{2+2\cos 60^o\cos 36^o}{2}\right]$$
$$\Rightarrow \dfrac{a^2}{2}\left[2+\not{2}\times \dfrac{1}{\not{2}}\cos 36^o\right]$$
$$d^2\Rightarrow \dfrac{a^2}{a}\left[2+\cos 36^o\right]$$
We know $$\cos 36=\dfrac{\sqrt{5}+1}{4}$$
$$\therefore d^2a^2=\dfrac{a^4}{2}\left[2+\dfrac{\sqrt{5}+1}{4}\right]=\dfrac{a^4}{2}\left[\dfrac{\sqrt{5}+9}{4}\right]$$
$$\Rightarrow$$ wrong options. [If d then]
Question 8
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The point $$P\left( x,y \right)$$ is equidistant from the points $$Q\left( c+d,d-c \right)$$ and $$R\left( c-d,c+d \right)$$ then

A
$$cx=dy$$

B
$$cx+dy=0$$

C
$$dx=cy$$

D
$$dx+cy=0$$

Solution

Point $$P(x,y)$$ is the equidistant from $$Q(c+d,d-x)$$ and $$R(c-d,c+d)$$
$$\therefore PQ=PR$$
$${ PQ }^{ 2 }={ PR }^{ 2 }$$
$${ \left[ x-\left( c+d \right) \right] }^{ 2 }+{ \left[ y-\left( d-c \right) \right] }^{ 2 }={ \left[ x-\left( c-d \right) \right] }^{ 2 }+{ \left[ y-\left( c+d \right) \right] }^{ 2 }$$
$${ x }^{ 2 }+{ \left( c+d \right) }^{ 2 }-2.x\left( c+d \right) +{ y }^{ 2 }+{ \left( d-c \right) }^{ 2 }-2.y(d-c)={ x }^{ 2 }+{ \left( c-d \right) }^{ 2 }-2.x.\left( c-d \right) +{ y }^{ 2 }+{ \left( c+d \right) }^{ 2 }+2.y(c+d)$$
Or $$x(c+d)+y(d-c)=x(c-d)+y(c+d)$$
Or $$(c+d)(x-y)=(c-d)(x+y)$$
Or $$cx-cy+dc-dy=cx+cy-dx-dy$$
Or $$2dx=2cy$$
Or $$dx=cy$$
$$\therefore $$Answer $$dx=cy$$
Question 9
1 / -0

The angle between the pair of lines whose equation is $$4{ x }^{ 2 }+10xy+m{ y }^{ 2 }+5x+10y=0$$ is

A
$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 8 } \right) }$$

B
$$\tan ^{ -1 }{ \dfrac {2 \sqrt { 25-4m } }{ m+4 } }$$

C
$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

D
$$\tan ^{ -1 }{ \dfrac { \sqrt { 25-4m } }{ m+4 } }$$

Solution

The angle b/w 2 line represented by $$ax^2+2hxy+by^2+2gx+2fy+c=0$$

$$\tan\theta=\Bigg|\dfrac{2\sqrt{h^2-ab}}{a+b}\Bigg|$$

Given the equation of line i.e. $$4x^2+10xy+my^2+5x+10y=0$$

Compare above equation with standard equation
$$a=4,h=5,b=m,g=\dfrac{5}{2},f=5,c=0$$

Hence $$\theta=\tan^{-1}\dfrac{2\sqrt{25-4m}}{m+4}$$
Question 10
1 / -0

Points $$A\left( 3,2,4 \right) ,B\left( \cfrac { 33 }{ 5 } ,\cfrac { 28 }{ 5 } ,\cfrac { 38 }{ 5 } \right) $$, and $$C(9,8,10)$$ are given. The ratio in which $$B$$ divides $$\overline { AC } $$ is

A
$$5:3$$

B
$$2:1$$

C
$$1:3$$

D
$$3:2$$

Solution

$$BA=\Bigg[[(\dfrac{33}{5})^2-(3)^2]+[(\dfrac{28}{5})^2-(2)^2]+[(\dfrac{38}{5})^2-(4)^2 ]\Bigg]=\Bigg(\sqrt{(\dfrac{18}{5})^2+(\dfrac{18}{5})^2+(\dfrac{18}{5})^2}\Bigg)=\sqrt{\dfrac{972}{25}}$$

$$CB=\Bigg[[9^2-(\dfrac{33}{5})^2]+[8^2-(\dfrac{28}{5})^2]+[10^2-(\dfrac{38}{5})^2 ]\Bigg]=\Bigg(\sqrt{(\dfrac{12}{5})^2+(\dfrac{12}{5})^2+(\dfrac{12}{5})^2}\Bigg)=\sqrt{\dfrac{432}{25}}
$$

Therefore $$\dfrac{BA}{CB}=\sqrt{\dfrac{972}{432}}=\dfrac32$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 21

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Straight Lines Test 21

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SHARING IS CARING

Question 1

$$-27$$ or $$8$$

$$-3$$ or $$2$$

$$-64$$ or $$27$$

$$-4$$ or $$3$$

Solution

Question 2

$$\sqrt{65}$$

$$\sqrt{74}$$

$$\sqrt{83}$$

$$\sqrt{97}$$

Solution

Question 3

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 4

$$a$$

$$\sqrt {b^{2} + c^{2}}$$

$$\sqrt {a^{2} + b^{2}}$$

$$a^{2} + b^{2}$$

Solution

Question 5

$$(1,6)$$

$$(-1/2,5)$$

$$(-7,1/6)$$

$$(5/6,6)$$

Solution

Question 6

One

Two

Three

Four

Solution

Question 7

$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +9 \right) }{ 4 } $$

$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 4 } $$

$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 8 } $$

$${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +1 \right) }{ 8 } $$

Solution

Question 8

$$cx=dy$$

$$cx+dy=0$$

$$dx=cy$$

$$dx+cy=0$$

Solution

Question 9

$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 8 } \right) }$$

$$\tan ^{ -1 }{ \dfrac {2 \sqrt { 25-4m } }{ m+4 } }$$

$$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $$

$$\tan ^{ -1 }{ \dfrac { \sqrt { 25-4m } }{ m+4 } }$$

Solution

Question 10

$$5:3$$

$$2:1$$

$$1:3$$

$$3:2$$

Solution

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