# Straight Line | Part-2 |Ex-2C

$1.~~A(4,6),~B(-1,3)~$ and $~C(2,-2)~$ are three given points . Find the length of the perpendicular from $~B~$ on $~AC.$

Solution.

Here the three given points are $~A(4,6),~B(-1,3)~$ and $~C(2,-2).$

So, the equation of $~AC~$ is

$\frac{y-6}{6-(-2)}=\frac{x-4}{4-2} \\ \text{or,}~~ \frac{y-6}{8}=\frac{x-4}{2} \\ \text{or,}~~ \frac 14(y-6)=(x-4) \\ \text{or,}~~ 4(x-4)=y-6 \\ \text{or,}~~ 4x-y-10=0 \rightarrow(1)$

Now, the length of the perpendicular from the point $~B(-1,3)~$ to the straight line $~(1)~$ is

$=\frac{|4(-1)-3-10|}{\sqrt{4^2+(-1)^2}}=\frac{|-4-3-10|}{\sqrt{16+1}}=\frac{17}{\sqrt{17}}=\sqrt{17}$

$2.~$ Find the perpendicular distance of the point $~(4,-1)~$ from the straight line through the points $~(1,1)~$ and $~(-11,-4).$

Solution.

The equation of the straight line joining the points $~(1,1)~$ and $~(-11,-4)~$ is

$\frac{y-1}{1-(-4)}=\frac{x-1}{1-(-11)} \\ \text{or,}~~ \frac{y-1}{5}=\frac{x-1}{12} \\ \text{or,}~~ 5(x-1)=12(y-1) \\ \text{or,}~~ 5x-12y-5+12=0 \\ \text{or,}~~ 5x-12y+7=0 \rightarrow(1)$

So, the perpendicular distance of the point $~(4,-1)~$ from the straight line $~(1)~$ is

$=\frac{|5 \times 4-12\times (-1)+7|}{\sqrt{5^2+(-12)^2}}=\frac{39}{13}=3~~\text{unit.}$

$3.~$ The equation to the base of an equilateral triangle is $~x+y=2~$ and its vertex is at $~(2,-1).$ Find the length of a side of the triangle.

Solution.

The equation to the base of an equilateral triangle is $~x+y=2~\rightarrow(1).$

Now, the perpendicular distance of the point $~(2,-1)~$ to the sl $~(1)~$ is

$=\frac{|2+(-1)-2|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}~~\text{unit}$

We know that the height of the equilateral triangle is

$~\frac{\sqrt{3}}{2} \times a~$ where $~a~$ is the length of the side of the $~\Delta$

$\therefore~ \frac{\sqrt{3}}{2} \times a=\frac{1}{\sqrt{2}} \\ \text{or,}~~a=\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}} =\frac{2}{\sqrt{6}}=\frac{2\sqrt{6}}{6}=\frac{\sqrt{6}}{3}~~\text{unit.}$

$4.~$ If the sum of the lengths of the perpendiculars dropped from a variable point $~P~$ on the two straight lines $~x+y=5~$ and $~3x-2y+7=0~$ be always equal to $~10~$ units, show that $~P~$ must move on a straight line.

Solution.

Let the co-ordinates of the variable point be $~P(h,k).$

Now, the distance of $~P(h,k)~$ to the straight line $~x+y=5~$ is

$=\frac{|h+k-5|}{\sqrt{1^2+1^2}}=\frac{|h+k-5|}{\sqrt{2}}~~\text{unit}$

Again, the distance of $~P(h,k)~$ to the straight line $~3x-2y+7=0~$ is

$=\frac{|3h-2k+7|}{\sqrt{3^2+(-2)^2}}=\frac{|3h-2k+7|}{\sqrt{13}}~~\text{unit}$

According to the problem,

$\frac{|h+k-5|}{\sqrt{2}}+\frac{|3h-2k+7|}{\sqrt{13}}=10 \\ \text{or,}~~ \pm \frac{h+k-5}{\sqrt{2}} \pm\frac{3h-2k+7}{\sqrt{13}}=10 \\ \text{or,}~~ \pm\sqrt{13}(h+k-5) \pm \sqrt{2}(3h-2k+7)\\~~=10\sqrt{26} \\ \text{or,}~~ (\pm \sqrt{13} \pm 3\sqrt{2})h+( \pm \sqrt{13}\\ \mp 2\sqrt{2})k \mp 5\sqrt{13}\pm 7\sqrt{2} -10\sqrt{26}=0$

So, the locus of the point $~P(h,k)~$ is

$(\pm \sqrt{13} \pm 3\sqrt{2})x+( \pm \sqrt{13} \mp 2\sqrt{2})y\\ \mp 5\sqrt{13}\pm 7\sqrt{2} -10\sqrt{26}=0.$

$5.~$ If $~p_1~$ and $~p_2~$ be the lengths of the perpendiculars from the origin upon the lines $~x\sin\theta+y\cos\theta=\frac a2 \sin2\theta~$ and $~x\cos\theta-y\sin\theta=a\cos2\theta~$,  prove that, $~4p_1^2+p_2^2=a^2.$

Solution.

The perpendicular distance of the straight line $~x\sin\theta+y\cos\theta=\frac a2 \sin2\theta~$  from the origin $~(0,0)~$ is given by

$~p_1=\frac{|0+0-(a/2)\sin2\theta|}{\sqrt{\sin^2\theta+\cos^2}}=|(a/2)\sin2\theta|$

Again, the distance of the straight line $~x\cos\theta-y\sin\theta=a\cos2\theta~$ from the origin $~(0,0)~$ is

$p_2=\frac{|0-0-a\cos2\theta|}{\sqrt{\cos^2\theta+\sin^2\theta}}=|a\cos2\theta|$

$\therefore~4p_1^2+p_2^2\\=4 \cdot \left(\frac a2 \sin2\theta\right)^2+(a\cos2\theta)^2\\=a^2(\sin^22\theta+\cos^22\theta)\\=a^2~~\text{(proved)}$

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$6.~$ Show that the product of two perpendiculars from the two points $~( \pm 4,0)~$ upon the sl $~3x\cos\theta+5y\sin\theta=15~$ is independent of $~\theta.$

Solution.

The perpendicular distance of the point $~(4,0)~$ from the sl $~3x\cos \theta+5y \sin \theta=15~$ is

$~d_1=\frac{|4 \times 3\cos p-15|}{\sqrt{(3\cos \theta)^2+(5\sin \theta)^2}}=\frac{3|4\cos \theta-5|}{\sqrt{9\cos^2\theta+25 \sin^2\theta}}$

Again, the perpendicular distance of the point $~(-4,0)~$ from the sl $~3x\cos\theta+5y\sin\theta=15~$ is

$d_2=\frac{|(-4) \times 3\cos p-15|}{\sqrt{(3\cos \theta)^2+(5\sin \theta)^2}}=\frac{3|4\cos \theta+5|}{\sqrt{9\cos^2\theta+25 \sin^2\theta}}$

Now, $~d_1d_2=\frac{9|(4\cos\theta-5)(4\cos\theta+5)|}{9\cos^2\theta+25\sin^2\theta}\\ \\ \text{or,}~ d_1d_2=\frac{9|16\cos^2\theta-25|}{9\cos^2\theta+25(1-\cos^2\theta)}=\frac{9(25-16\cos^2\theta)}{25-16\cos^2\theta}\\~~[\because~ \cos^2\theta \leq 1\Rightarrow 25>16 \geq 16\cos^2\theta] \\ \therefore~ d_1d_2=9~~\text{(constant)}$

Hence, we can conclude that the product of two perpendiculars  is independent of $~\theta.$

$7.~$ Find the equations of the straight lines through $~(0,a)~$ on which the perpendiculars dropped from the point $~(2a,2a)~$ are each of length $~a~$ unit.

Solution.

The equation of the straight line passing through the point $~(0,a)~$ and having slope $~m~$ can be written as

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

Now, the perpendicular distance of the point $~(2a,2a)~$ from the straight line $~(1)~$ is $~a~$ unit.

$\therefore~ \frac{|m \times 2a-2a+a|}{\sqrt{m^2+1}}=a \\ \text{or,}~~ \frac{|2ma-a|}{\sqrt{m^2+1}}=a \\ \text{or,}~~|a(2m-1)|=a\sqrt{m^2+1} \\ \text{or,}~~ a^2(2m-1)^2=a^2(m^2+1) \\ \text{or,}~~ (2m-1)^2=m^2+1 \\ \text{or,}~~ 4m^2-4m+1=m^2+1 \\ \text{or,}~~ 3m^2-4m=0 \\ \text{or,}~~ m(3m-4)=0 \\ \therefore~ m=0, \frac 43.$

Hence , by the equation $~(1)~$ , we get the required straight line is (for $~m=0~$)

$~0-y+a=0 \Rightarrow y=a$ and

for $~m=4/3,~$ the required straight line  is

$~\frac 43x-y+a=0 \Rightarrow 4x-3y+3a=0.$

$8.~$ Find the equation of a straight line mid-way between the lines $~2x+3y=5~$ and $~2x+3y+1=0.$

Solution.

The equation of the given straight lines are $~2x+3y-5=0 \rightarrow(1)~$ and $~2x+3y+1=0 \rightarrow(2).$

Since the required straight line must be parallel to both straight lines $~(1)~$ and $~(2)~$ and so suppose that the straight line be $~2x+3y-k=0~(k \neq 0)\rightarrow(3).$

Since the straight line $~(3)~$ is mid-way between the straight lines $~(1)~$ and $~(2)~$, so

$\frac{|-k-1|}{\sqrt{2^2+3^2}}=\frac{|-k-(-5)|}{\sqrt{2^2+3^2}} \\ \text{or,}~~ \frac{-k-1}{\sqrt{4+9}}=\pm \frac{-k+5}{\sqrt{4+9}}\rightarrow(4) \\ \text{or,}~~ -k-1=\pm(-k+5) \\ \therefore~ -k-1=-(-k+5) \\ \text{or,}~~5-1=2k \\ \text{or,}~~ k=\frac 42=2$

By taking +ve sign , we get from $~(4),~~~-1=5~$ which is impossible.

So, the required equation of straight line is

$~2x+3y-2=0~~[\text{By (3)}].$