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. $
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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 $~(4),$
$~7 \times 1-4 \times 1+k''=0 \Rightarrow k''=-3.$
$\therefore~ AD \equiv 7x-4y-3=0.$
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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 $~(2-\sqrt{3})x-y-2(2-\sqrt{3})+3=0 \\ \text{or,}~~(2-\sqrt{3})x-y=1-2\sqrt{3}.$
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