Straight Lines Test -2

Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.

Hence option 1 is correct.

The distance of a  point from X axis is its y coordinate.

Hence the distance of the point (α,β) from X axis  is |β|

Let L be the foot of the perpendicular from the X axis. Therefore its y coocrdinate is zero

Therefore the coordiantes of the point L is (x,0)

Hence option 1 is the correct answer

Since the angle made with x axis is zero, since tanθθ is the slope. Then tan0 = 0

Hence the slpoe of the line parallel to X-axis is zero

Option 1 is the correct answer.

The lines formed by these lines is right angled, triangle.

Two lines are perpendicular to each other if the product  od their slopes is -1

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

The product of these slopes is -1

Hence the lines3x - y – 4 = 0 , x + 3y – 4 = 0  are perpendicular to each other.

Therefore the triangle formed by these lines is a right angled triangle.

Given lines are,
$$r[2 \cos \theta  + 5\sin\theta ] = 3$$ and
$$r[2 \sin \theta  - 5\cos\theta ]+ 4 = 0$$
Cartesian form is,
$$2x+5y=3$$
$$2y-5x=-4$$
Slopes of these lines are $$\displaystyle -\frac{2}{5}$$ and $$\displaystyle\frac {5}{2}$$
Here, product of slopes $$=-1$$
Thus, the lines are perpendicular.

For $$({ a }+{ b }){ x }+({ a }-{ b }){ y }-2{ a }{ b }=0$$
Slope is $${ m }_{ 1 }=-\cfrac { a+b }{ a-b } $$
And for $$({ a }-{ b }){ x }+({ a }+{ b }){ y }-2{ a }{ b }=0$$
Slope is $${ m }_{ 2 }=-\cfrac { a-b }{ a+b } $$
Then angle is
$$\tan { \theta  } =\left| \cfrac { { m }_{ 1 }-{ m }_{ 2} }{ 1+{ m }_{ 1 }{ m }_{ 2 } }  \right| =\left| \cfrac { -\cfrac { a+b }{ a-b } +\cfrac { a-b }{ a+b }  }{ 1+\cfrac { a+b }{ a-b } \cfrac { a-b }{ a+b }  }  \right| \\ =\left| \cfrac { { a }^{ 2 }+{ b }^{ 2 }-2ab-{ a }^{ 2 }-{ b }^{ 2 }-2ab }{ { a }^{ 2 }-{ b }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 } }  \right| =\left| \cfrac { 2ab }{ { a }^{ 2 }-{ b }^{ 2 } }  \right| $$

$$x \cos \alpha+y \sin \alpha=p_1$$

$$slope=m_1=\dfrac{-\cos  \alpha}{\sin \alpha}=-\cot\alpha$$

And $$x \cos \beta+y \sin  \beta=p_2$$

$$slope =m_2=\dfrac{-\cos \beta}{\sin \beta}=-\cot\beta$$

Therefore, 

$$\tan\theta=\left | \dfrac{-\cot \alpha+\cot\beta}{1+\cot\alpha cot\beta} \right |$$

$$\tan\theta=\left | \dfrac{\tan\alpha- \tan\beta}{1+\tan\alpha \tan\beta} \right |$$

$$\tan\theta=|\tan (\alpha-\beta)|$$

$$\Rightarrow \theta=\alpha-\beta$$

Line passing through points $$(2,\sin\theta)$$ and $$(1,\cos\theta)$$ is
$$y-\cos\theta=(\sin\theta-\cos\theta)(x-1)$$
$$y=(\sin\theta-\cos\theta)x+2\cos\theta-sin\theta$$
Slope$$=\sin\theta-\cos\theta=0$$   (given)
$$\sin\theta=\cos\theta$$
$$\tan\theta=1$$
$$\theta=n\pi+\dfrac{\pi}{4},\forall n\in Z$$

Let $$m$$ be slope of the line through A

$$kx-y+6=0$$

$$slope=m_1= \dfrac{-k}{-1}=k$$

And $$3x-5y+7=0$$

$$slope=m_2= \dfrac{-3}{-5}= \dfrac{3}{5}$$

So, angle between this two lines is $$ \dfrac{\pi}{4}$$

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

$$1=\left |  \dfrac{k- \dfrac{3}{5}}{1+ \dfrac{3k}{5}} \right |$$

$$\therefore 1+ \dfrac{3k}{5}=k- \dfrac{3}{5}$$

$$ \dfrac{8}{5}= \dfrac{2k}{5}$$

$$k=4$$

$$kx+y+9=0$$
$$y=-kx-9$$
Thus, slope of the line $$m_{1}=-k$$

Similarly, $$y=3x+4$$
Slope of the line is $$m_{2}=3$$
Therefore, angle between the lines is

$$\tan\theta=\left|\dfrac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|$$

$$=\left|\dfrac{3+k}{1-3k}\right|$$

$$=\left|\dfrac{k+3}{1-3k}\right|$$

But $$\theta=45^{o}$$
 
$$\therefore \tan\theta=1$$

$$\Rightarrow \dfrac{k+3}{1-3k}=\pm1$$

$$\Rightarrow k+3=\pm(1-3k)$$
Case I
$$k+3=1-3k$$
i.e. $$4k=-2$$
$$\Rightarrow k=\dfrac{-1}{2}$$

Case II
$$k+3=-1+3k$$

$$\Rightarrow -2k=-4$$

$$\therefore k=2$$
Hence, $$k=2,\dfrac{-1}{2}$$

Given  $$r \cos (\theta -\alpha )= p$$.......(i)
$$ r \sin (\theta -\alpha )= q $$    ...(ii)
Now, equation of line perpendicular to given line is 
$$r\sin (\dfrac{\pi}{2}+ (\theta -\alpha ))=k$$
$$\Rightarrow r \cos (\theta -\alpha )=k$$
which is of the form of given equation (i)

Let $$A(2,3), B(6,7)$$ and $$C(8,3)$$ be the vertices of parallelogra

The  given  points  are  $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,1 \right) ,  B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 12,2 \right)   \&   C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,2 \right)$$

Let  us  obtain  $$ar\Delta ABC$$  by  applying  the  area  formula.

$$ ar\Delta =\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right)  \right\}$$

$$\Rightarrow ar\Delta =\dfrac { 1 }{ 2 } \left\{ 3\left( 2-2 \right) +12\left( 2-1 \right) +0\left( 1-2 \right)  \right\}$$ units $$=6$$ units

$$\therefore   ar\Delta =6$$ units 

Hence, option B.