Ad-1

if( aicp_can_see_ads() ) {

Straight Line | Part-6 | Ex-2B

Straight Line | Part-6 |Ex-2B

 
Here are some interesting facts about straight lines in mathematics:

1. A straight line is the shortest distance between two points: Among all possible curves, a straight line provides the shortest path between any two given points. This property is known as the "straight line segment theorem" and is fundamental in geometry.

2. A straight line has a slope but no curvature: Unlike curves, straight lines do not exhibit any curvature. Their slope remains constant, indicating a consistent rate of change along their entire length.

3. Parallel lines never intersect: If two lines have the same slope and are in the same plane, they are considered parallel. Parallel lines extend infinitely and never cross each other, regardless of their distance.

4. Every straight line has a perpendicular line: Given a straight line, there exists a unique line that intersects it at a right angle, forming a 90-degree angle. This line is called the perpendicular line and is useful in geometric constructions and calculations.

5. The equation of a straight line can determine its properties: The equation y = mx + b, also known as the slope-intercept form, provides essential information about a straight line. The slope (m) determines the line's steepness, while the y-intercept (b) indicates where the line intersects the y-axis.

6. Straight lines can be described using different forms: Apart from the slope-intercept form, straight lines can also be represented in other forms, such as the point-slope form (y - y₁ = m(x - x₁)) and the general form (Ax + By = C), where A, B, and C are constants.

7. Three non-collinear points uniquely determine a straight line: If three points in a plane are not on the same straight line (non-collinear), they uniquely determine a straight line. This property is utilized in coordinate geometry and determining the equations of lines.

These facts highlight some intriguing aspects of straight lines in mathematics and demonstrate their significance in various mathematical concepts and real-world applications.


11.  Show that the lines $~(a+b)x+(a-b)y-2ab=0,~(a-b)x+(a+b)y-2ab=0~$ and $~x+y=0~$ form an isosceles triangle whose vertical angle is $~2\tan^{-1}(a/b).$


Solution.


The given lines are 

$~(a+b)x+(a-b)y-2ab=0 \rightarrow(1),\\~(a-b)x+(a+b)y-2ab=0 \rightarrow(2),\\~ x+y=0\rightarrow(3) $

On comparing the aforesaid equations with $~y=mx+c~,$ we get the slopes of $~(1),~(2),~(3)~$ which are $~(m_1)=-\left(\frac{a+b}{a-b}\right),\\~ (m_2)=-\left(\frac{a-b}{a+b}\right),~(m_3)=-1.$

If $~\alpha~$ is the angle between $~(1)~$ and $~(3)~$ , then it is given by 

$\tan\alpha =\left|\frac{m_1-m_3}{1+m_1m_3}\right|=\left|\frac{-\frac{a+b}{a-b}+1}{1+\frac{a+b}{a-b}}\right| \\ \text{or,}~~\tan\alpha=\left|\frac{-(a+b)+a-b}{a-b+a+b}\right| \\ \text{or,}~~\tan\alpha=\left|\frac{-2b}{2a}\right|=\frac ba \rightarrow(4)$

If $~\beta~$ is the angle between $~(2)~$ and $~(3)~$ , then it is given by 

$\tan\beta =\left|\frac{m_2-m_3}{1+m_2m_3}\right|=\left|\frac{-\frac{a-b}{a+b}+1}{1+\frac{a-b}{a+b}}\right| \\ \text{or,}~~\tan\beta=\left|\frac{-(a-b)+a+b}{a+b+a-b}\right| \\ \text{or,}~~\tan\beta=\left|\frac{2b}{2a}\right|=\frac ba \rightarrow(5)$

So, by $~(4)~$ and $~(5)~$ we get,

$\tan\alpha=\tan\beta \\ \therefore~ \beta=\alpha=\tan^{-1}(b/a).$

Hence, the triangle is an isosceles triangle with vertical angle $~~\pi-2\alpha=\pi-2\tan^{-1}(b/a).$


12.  Find the equations of the straight lines which pass through $~(3,2)~$ and make an angle of $~45^{\circ}~$ with the line $~x=2y+4.$


Solution.


The equation of any straight line having slope $~m~$ and passing through the point $~(3,2)~$ can be written as 

$~y-2=m(x-3) \\ \text{or,}~~ mx-y-3m+2=0 \rightarrow(1)$

Also, the given straight line is $~x-2y-4=0\rightarrow(2)$. 

Slope of the straight line $~(2)~$ is $~(m')=\frac 12.$

If $~45^{\circ}~$ is the angle between the straight lines $~(1)~$ and $~(2)~$, then 

$\tan 45^{\circ}=\left|\frac{m-m'}{1+mm'}\right|=\left|\frac{m-\frac 12}{1+m \cdot \frac 12}\right| \\ \text{or,}~~ 1=\left|\frac{2m-1}{2+m}\right| \\ \text{or,}~~ \frac{2m-1}{2+m}=\pm 1 \\ \text{or,}~~ 2m-1=\pm(2+m) \\ \therefore~ 2m-1=2+m \Rightarrow m=3,\\~~~ 2m-1=-2-m \Rightarrow m=-\frac 13.$

Hence, by $~(1)~$ we get for $~m=3,$

$3x-y-3 \times 3+2=0 \Rightarrow 3x-y=7.$

Again, by $~(1)~$ we get for $~m=-1/3,$

$-\frac 13x-y-3 \times (-1/3)+2=0 \\ \text{or,}~~ -\frac x3-y+3=0 \\ \text{or,}~~  -\frac x3-y=-3 \\ \text{or,}~~  \frac x3+y=3 \\ \therefore~ x+3y=9.$

Hence, the equations of the straight lines  are $~3x-y=7~$ and $~x+3y=9.$


13.  Find the equations of the straight lines which pass through the origin and make an angle of $~75^{\circ}~$ with the line $~x+y+\sqrt{3}(y-x)=0$


Solution.


The given sl can be written as $~x(1-\sqrt{3})+y(1+\sqrt{3})=0 \rightarrow(1)$

The slope of straight line $~(1)~$ is $~(m_1)=-\frac{1-\sqrt{3}}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ 

Let the equations of the other straight line which pass through the origin and makes an angle of $~75^{\circ}~$ with the given line is $~y=m_2x.$

$\therefore~ \tan 75^{\circ}=\left|\frac{m_2-m_1}{1+m_1m_2}\right| \\ \text{or,}~~ \tan 75^{\circ}=\left|\frac{m_2-\frac{\sqrt{3}-1}{\sqrt{3}+1}}{1+m_2 \cdot \frac{\sqrt{3}-1}{\sqrt{3}+1}}\right| \\ \text{or,}~~ \mp \tan 75^{\circ}=\frac{m_2(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+m_2(\sqrt{3}-1)}~~[*] \\ \text{or,}~~ -(2+\sqrt{3})=\frac{m_2(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+m_2(\sqrt{3}-1)} \\~~[\text{taking -ve sign}] \\ \text{or,}~~ (2+\sqrt{3})=\frac{(\sqrt{3}-1)-m_2(\sqrt{3}+1)}{(\sqrt{3}+1)+m_2(\sqrt{3}-1)} \\ \text{or,}~~ (2+\sqrt{3})(\sqrt{3}+1)+m_2(2+\sqrt{3})(\sqrt{3}-1)\\=(\sqrt{3}-1)-m_2(\sqrt{3}+1) \\ \text{or,}~~ (2+3+3\sqrt{3})+m_2(2\sqrt{3}-2+3-\sqrt{3})\\=(\sqrt{3}-1)-m_2(\sqrt{3}+1) \\ \text{or,}~~ 6+2\sqrt{3}=-2m_2(\sqrt{3}+1) \\ \text{or,}~~ 2\sqrt{3}(\sqrt{3}+1)=-2m_2(\sqrt{3}+1) \\ \therefore~ m_2=-\sqrt{3}$

Again, from $~(*)~$, by taking +ve values of $~\tan 75^{\circ}~$, we get the value of $~m_2=\infty.$

Hence, the equation of the line 

$~y=m_2x \Rightarrow y=-\sqrt{3}x \Rightarrow y+\sqrt{3}x=0$

Again, for $~m_2=\infty,~$ the equation of the line is 

$~y=m_2x \Rightarrow x=\frac{1}{m_2}y=0 \times y=0. $


Oswaal CBSE Chapterwise Solved Papers 2023-2014 Applied Mathematics Class 12th (2024 Exam) 



14.  Two straight lines pass through the point $~(-2,5)~$ such that one of them makes an angle of $~\tan^{-1}\frac 34~$ with the given line $~x-y+5=0~$ and the given line makes an angle of $~\tan^{-1}\frac 23~$ with the other line. Find the equations to the two lines.


Solution.



Any equation of straight line passing through $~(-2,5)~$ and having  slope $~m~$ can be written as $~y-5=m[x-(-2)]\\ \Rightarrow y-5=m(x+2)\rightarrow(1)$


Also, given straight line is $~x-y+5=0~$ having slope $~1.$


$\therefore~ \tan^{-1} \frac 34=\tan^{-1} \left|\frac{1-m}{1+m}\right| \\ \text{or,}~~ \frac 34=\pm\frac{1-m}{1+m} \\ \text{or,}~~ 3(1+m)=\pm 4(1-m) \\ \therefore 3(1+m)=4(1-m)\Rightarrow m=\frac 17,\\~3(1+m)=-4(1-m)\Rightarrow m=7.$


For $~m=\frac 17 ~$, we get by $~(1),$


$~y-5=\frac 17(x+2)\\ \text{or,}~~ 7(y-5)=x+2 \\ \text{or,}~~ 7y-35=x+2\\ \text{or,}~~ x-7y+37=0$


For $~m=7 ~$, we get by $~(1),$


$~y-5=7(x+2)\\ \text{or,}~~ y-5=7x+14 \\ \text{or,}~~ 7x-y+19=0$


Any equation of straight line passing through $~(-2,5)~$ and having  slope $~m'~$ can be written as $~y-5=m'[x-(-2)] \\ \Rightarrow y-5=m'(x+2)\rightarrow(2)$


$\therefore~ \tan^{-1} \frac 23=\tan^{-1} \left|\frac{1-m'}{1+m'}\right| \\ \text{or,}~~ \frac 23=\pm\frac{1-m'}{1+m'} \\ \text{or,}~~ 2(1+m')=\pm 3(1-m') \\ \therefore 2(1+m')=3(1-m')\Rightarrow m'=\frac 15,\\~2(1+m')=-3(1-m')\Rightarrow m'=5.$


For $~m'=\frac 15 ~$, we get by $~(2),$ the equation of straight line


 $~ x-5y+27=0.$


For $~m'=5 ~$, we get by $~(2),$ the equation of straight line


 $~ 5x-y+15=0.$


15.  The equation of a side of a rectangle is $~4x+7y+5=0~$ and the two vertices are $~(-3,1)~$ and $~(1,1).~$ Find  the equations of the other three sides.


Solution.




The straight line $~4x+7y+5=0~$ passes through the point $~(-3,1)~$ and so,

$~AB \equiv 4x+7y+5=0\rightarrow(1).$

Now, $~BC \perp AB~$  so, any straight line perpendicular to $~(1)~$ can be written as $~7x-4y+k=0~( k\neq 0) \rightarrow(2).$

Since the straight line $~(2)~$ passes through the point $~B(-3,1),$

$\therefore~ 7 \times (-3)-4 \times 1+k=0 \Rightarrow k=25.$

So,$~BC \equiv 7x-4y+25=0.$

Again, $~ CD~ ||~ AB~$ and so,

$~CD \equiv 4x+7y+k'=0\rightarrow(3)$

Now, since the straight line $~CD~$ passes through the point $~D(1,1)~,$ we get by $~(3),~$ 

$~4 \times 1+7 \times 1+k'=0 \Rightarrow k'=-11.$

Hence, $~ CD \equiv 4x+7y-11=0.$

Again, since $~AD~||~BC~$ , so, $~AD \equiv 7x-4y+k''=0\rightarrow(4).$

Since the straight line $~(4)~$ passes through the point $~D(1,1)~$,we get by $~(4),$

$~7 \times 1-4 \times 1+k''=0 \Rightarrow k''=-3.$ 

$\therefore~ AD \equiv 7x-4y-3=0.$


Career Crush Paperback 




16.  Find the distance of the line $~3x-2y-1=0~$ from the point $~(5,3)~$ measured in a direction making an angle $~45^{\circ}~$ with the given line.


Solution.




Let $~3x-2y-1=0 \rightarrow(1)$

Now, Suppose that the distance of the point $~D(5,3)~$ from the straight line $~(1)~$ having inclination of $~45^{\circ}~$ (as in the picture) is $~k~$ unit.

Now, the perpendicular distance of $~(1)~$ from the point $~D~$ is 

$~d=\frac{3 \times 5-2\times 3-1}{\sqrt{3^2+(-2)^2}}=\frac{8}{\sqrt{13}}.$

Clearly, $~\sin 45^{\circ}=\frac dk \\ \text{or,}~~ \frac{1}{\sqrt{2}}=\frac{8}{k\sqrt{13}} \\ \text{or,}~~ k=\frac{8\sqrt{26}}{13}.$

So, the required distance $=\frac{8\sqrt{26}}{13}~\text{unit.}$


17.  The co-ordinates of one vertex of an equilateral triangle are $~(2,3)~$ and the equation of the opposite side is $~x+y=2.~$ Find the equations of its other sides.


Solution.




Other two sides of the triangle passes through the point $~(2,3)~$ .Now, the equation of any straight line passing through the point $~(2,3)~$ and having slope $~m,~$ can be written as 

$~y-3=m(x-2) \Rightarrow mx-y-2m+3=0\rightarrow(1)$

$~\because~$ the triangle is  equilateral, so

$~\tan 60^{\circ}=\left|\frac{m+1}{m-1}\right| \\ \text{or,}~~\pm\sqrt{3}= \left(\frac{m+1}{m-1}\right) \\ \therefore \frac{m+1}{m-1}=\sqrt{3}\rightarrow(2),\\~~\frac{m+1}{m-1}=-\sqrt{3}\rightarrow(3)$

From $~(2)~$ we get,

$~\frac{m+1}{m-1}=\sqrt{3} \\ \text{or,}~~\frac{(m+1)+(m-1)}{(m+1)-(m-1)}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\ \therefore m=\frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2-1^2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}.$

Similarly, by $~(3)~$ we get, $~m=2-\sqrt{3}.$

Hence using $~(1)~$ we get the required equations of straight lines which are

$~(2+\sqrt{3})x-y-2(2+\sqrt{3})+3=0 \\ \text{or,}~~(2+\sqrt{3})x-y=1+2\sqrt{3}$

and $~(2-\sqrt{3})x-y-2(2-\sqrt{3})+3=0 \\ \text{or,}~~(2-\sqrt{3})x-y=1-2\sqrt{3}.$ 


Post a Comment

0 Comments
* Please Don't Spam Here. All the Comments are Reviewed by Admin.