Differentiation (Part-38) | S N De

 

1.  The equations of the sides $~AB,~BC~$ and $~CA~$ of the $~\Delta ABC~$ are $~3x+4y+3=0,~2x+y+1=0~$ and $~2x+3y+1=0~$ respectively. Find the equation of its altitude through the vertex $~A.$

Solution.

We have,

$~AB \equiv 3x+4y+3=0\rightarrow(1),\\~ BC \equiv 2x+y+1=0\rightarrow(2),\\~ CA \equiv 2x+3y+1=0\rightarrow(3).$

From $~(1)~$ and $~(3)~$ we get,

$~2(3x+4y+3)-3(2x+3y+1)=0  \\ \text{or,}~~8y+6-9y-3=0 \\ \text{or,}~~ -y+3=0 \\ \therefore~y=3.$

Putting the value of $~y~$ in $~(1)~$, we get

$3x+ 4 \times 3+3=0 \\ \text{or,}~~ 3x=-15 \\ \therefore x=\frac{-15}{3}=-5.$

Hence, solving $~(1)~$ and $~(3)~$,we get, $~A\equiv(-5,3).$

Now, the equation of any straight line perpendicular to $~BC~$ is :

$~x-2y+k=0\rightarrow(4)$

Since the straight line $~(4)~$ passes through the point $~A(-5,3)~$, so

$-5-2 \times 3+k=0 \Rightarrow k=11\rightarrow(5).$

So, finally by $~(4)~$ and $~(5)~$, we get the required equation of its altitude through the vertex $~A~$ which is $~x-2y+11=0.$

2. The equations of two sides of a triangle are $~x+4y=7~$ and $~2x-5y=1~$. If the equations of its base be $~x+y=2,~$ find the length and the equation of its altitude.

Solution.

Let the triangle be $~\Delta ABC~$ where 

$~AB \equiv x+4y=7 \rightarrow(1),\\~AC \equiv 2x-5y=1\rightarrow(2),\\~ BC \equiv x+y=2 \rightarrow(3).$

From $~(1)~$ we get, $~x=7-4y \rightarrow(4).$

From $~(2)~$ and $~(4)~$ we get,

$~2(7-4y)-5y=1 \\ \text{or,}~~ 14-8y-5y=1 \\ \text{or,}~~ -13y=1-14 \\ \text{or,}~~ y=\frac{-13}{-13}=1,\\~\therefore~ x=7-4\times 1=3~~[\text{By (4)}]$

$\therefore~ A \equiv (3,1).$

Now, the length of altitude of $~\Delta ABC~$

$=\text{distance of}~A~ \text{from}~ BC\\=\left|\frac{3+1-2}{\sqrt{1^2+1^2}}\right|\\=\left|\frac{2}{\sqrt{2}}\right|\\=\sqrt{2}~~\text{unit.}$ 

Now, the equation of any straight line perpendicular to $~BC~$ can be  written as  $~x-y+k=0~~(k \neq 0)\rightarrow(5).$

Since the straight line passes through $~A(3,1)~$ , so

$~3-1+k=0 \Rightarrow k=-2.$

Hence, the equation of any straight line perpendicular to $~BC~$ is $~x-y-2=0.$

So, the equation of its altitude is $~x-y=2.$

3. Show that the product of the perpendiculars drawn from the two points $~(\pm \sqrt{a^2-b^2},0)~$ upon the sl $~\frac xa \cos\theta+\frac yb \sin\theta=1~$ is $~b^2.$

Solution.

We have the equation of the sl $~\frac xa \cos\theta+\frac yb \sin\theta=1~\rightarrow(1)$ The perpendicular distance of $~(1),~$ from the point $~(\sqrt{a^2-b^2},0)$ $=\frac{\left|\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2}+\frac{\sin^2\theta}{b^2}}}\\=\frac{|b\sqrt{a^2-b^2}\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$ Similarly, the perpendicular distance of $~(1),~$ from the point $~(-\sqrt{a^2-b^2},0)$ $=\frac{\left|-\frac{\sqrt{a^2-b^2}}{a}\cos\theta-1\right|}{\sqrt{\frac{\cos^2 \theta}{a^2}+\frac{\sin^2\theta}{b^2}}}\\=\frac{|b\sqrt{a^2-b^2}\cos\theta+ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$ $\therefore~$ the product of the perpendiculars drawn from the two points $~(\pm \sqrt{a^2-b^2},0)~$ is $=\frac{|b\sqrt{a^2-b^2}\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}} \\~~\times \frac{|b\sqrt{a^2-b^2}\cos\theta+ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}\\=\frac{|b^2(a^2-b^2)\cos^2\theta-a^2b^2|}{b^2\cos^2\theta+a^2\sin^2\theta}\\=\frac{|-a^2b^2(1-\cos^2\theta)-b^4\cos^2\theta|}{b^2\cos^2\theta+a^2\sin^2\theta}\\=\frac{|-a^2b^2\sin^2\theta-b^4\cos^2\theta|}{b^2\cos^2\theta+a^2\sin^2\theta} \\=\frac{|-b^2(b^2\cos^2\theta+a^2\sin^2\theta)|}{b^2\cos^2\theta+a^2\sin^2\theta}\\=|-b^2|\\=b^2.$

4. One side of an equilateral triangle is the line $~5y=12x-3~$ and its centroid is at $~(2,-1);~$ find the length of a side of the triangle. 

Solution.

The distance of the given sl $~12x-5y-3=0 \rightarrow(1)~$ from the point $~(2,-1)~$ is 

$=\left|\frac{12 \times 2-5\times (-1)-3}{\sqrt{12^2+(-5)^2}}\right|\\=\left|\frac{24+5-3}{\sqrt{169}}\right|\\=\left|\frac{26}{13}\right|\\=2~~\text{unit}.$

Since the given triangle is equilateral, so its centroid and orthocentre coincide.

Let $~a~$ be the length of the side of the triangle so that the height of the triangle is given by 

$\frac{\sqrt{3}}{2} \times a=2 \times 3 \\ \text{or,}~~ a=4\sqrt{3}.$

Hence, the length of a side of the triangle is $~4\sqrt{3}~$ unit.

5. Find the image of the point $~(-3,-1)~$ with respect to the straight line $~3y+2x+22=0.$

Solution.

The equation of straight line perpendicular to $~2x+3y+22=0 \rightarrow(1)~$ can be written as $~3x-2y+k=0~~(k\neq 0)\rightarrow(2)$

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

$~ 3\times (-3)-2 \times (-1)+k \Rightarrow k=7.$

So, the straight line $~(2)~$ can be written as $~3x-2y+7=0 \rightarrow(3).$

From $~(1)~$ and $~(3)~$ we get,

$~2(3x-2y+7)-3(2x+3y+22)=0 \\ \text{or,}~~ -13y+14-66=0 \\ \text{or,}~~ -13y=52 \\ \therefore~ y=\frac{52}{-13}=-4.$

Putting the value of $~y~$ in $~(3)~$ we get,

$~ 3x-2 \times (-4)+7=0 \\ \text{or,}~~ 3x+15=0 \\ \therefore~ x=\frac{-15}{3}=-5.$

So, the point of intersection of the straight lines $~(1)~$ and $~(3)~$ is $~(-5,-4).$

Let the image of the point $~(-3,-1)~$ be $~(a,b).$

$\therefore~\frac{a-3}{2}=-5\Rightarrow a=-7,\\~\frac{b-1}{2}=-4 \Rightarrow b=-7.$

Hence, the image of the point $~(-3,-1)~$ with respect to the straight line $~3y+2x+22=0~$  is $~(-7,-7).$

6. Show that any point on the straight line $~11x-3y+11=0~$ is equidistant from the straight lines $~12x+5y+12=0~$ and $~3x-4y+3=0.$

Solution.

Suppose that the  point $~P(h,k)~$ is equidistant from the given two straight lines.

$\therefore~\frac{12h+5k+12}{\sqrt{12^2+5^2}}=\pm\frac{3h-4k+3}{\sqrt{3^2+(-4)^2}} \\ \text{or,}~~ \frac{12h+5k+12}{13}=\pm \frac{3h-4k+3}{5} \rightarrow(1)$

Now, taking -ve sign, we get from $~(1),$

$~5(12h+5k+12)=-13(3h-4k+3) \\ \text{or,}~~ 60h+39h+25k-52k+60+39=0 \\ \text{or,}~~ 99h-27k+99=0 \\ \text{or,}~~ 9(11h-3k+11)=0 \\ \therefore~ 11h-3k+11=0\rightarrow(2).$

Now, by $~(2)~$ we can clearly say that the point $~(h,k)~$ lies on the straight line $~11x-3y+11=0.$

Hence, we can finally say that any point on the straight line $~11x-3y+11=0~$ is equidistant from the straight lines $~12x+5y+12=0~$ and $~3x-4y+3=0.$

7. Find the equation of the straight line which is perpendicular to the sl $~3x-2y+5=0~$ and whose distance from the origin is equal to the perpendicular distance of the given line from the point $~(2,-1).$

Solution.

The equation of any straight line perpendicular to $~3x-2y+5=0~$ can be written as $~2x+3y+k=0~~(k \neq 0)~\rightarrow(1).$

Now, the distance of the straight line $~(1)~$ from the origin 

$\left|\frac{2 \times 0+3 \times 0+k}{\sqrt{2^2+3^2}}\right|=\left|\frac{k}{\sqrt{13}}\right|~\rightarrow(2)$

Again, the distance of the given straight line  from the point $~(2,-1)~$ is 

$\left|\frac{3 \times 2-2\times (-1)+5}{\sqrt{3^2+(-2)^2}}\right|=\left|\frac{13}{\sqrt{13}}\right|\rightarrow(3)$

So, from $~(2)~$ and $~(3)~$ we get,

$\left|\frac{k}{\sqrt{13}}\right|=\left|\frac{13}{\sqrt{13}}\right| \Rightarrow k=\pm 13.$

Hence, the equation of the required straight line is $~2x+3y\pm 13=0$.

8. Show that the perpendicular dropped from any point of the straight line $~9x+3y=20~$ upon the two straight lines $~x+3y=6~$ and $~13x-9y=10~$ are equal.

Solution.

Suppose that the perpendicular distance of the point $~P(h,k)~$ to the two given lines $~x+3y=6~$ and $~13x-9y=10~$ are equal.

$\therefore~ \frac{|h+3k-6|}{\sqrt{1^2+3^2}}=\frac{13h-9k-10}{\sqrt{13^2+(-9)^2}} \\ \text{or,}~~ \frac{h+3k-6}{\sqrt{10}}=\pm \frac{13h-9k-10}{\sqrt{250}} \\ \text{or,}~~ \frac{h+3k-6}{\sqrt{10}}=\pm \frac{13h-9k-10}{5\sqrt{10}} \\ \text{or,}~~ h+3k-6=\pm \frac 15(13h-9k-10) \\ \text{or,}~~ 5(h+3k-6)=\pm(13h-9k-10)\rightarrow(1)$

Taking -ve sign, we get from $~(1),$

$~5h+15k-30=-(13h-9k-10) \\ \text{or,}~~ 5h+13h+15k-9k-30-10=0 \\ \text{or,}~~ 18h+6k-40=0 \\ \text{or,}~~ 2(9h+3k-20)=0 \\ \therefore~ 9h+3k-20=0\rightarrow(2)$

Hence, by $~(2),~$ we can conclude that the point $~P(h,k)~$ lies on the straight line $~9x+3y=20.$

9. Find the foot of the perpendicular from the point $~(-2,6)~$ on the sl $~2x+3y=1.~$ What are the co-ordinates of the point which is the image of $~(-2,6)~$ with respect to the given line.

Solution.

Equation of any straight line perpendicular to $~2x+3y=1\rightarrow(1)~$ is $~3x-2y+k=0~(k \neq 0)\rightarrow(2)$ . Since the straight line $~(2)~$ passes through the point $~(-2,6),$

$3 \times (-2)-2 \times 6+k=0 \Rightarrow k=18$

So, the required equation of straight line perpendicular to $~(2)~$, is 

$~3x-2y+18=0~~[\text{By (2)}]\rightarrow(3)$

Now, from $~(1)~$ and $~(3)~$ we get,

$~3(2x+3y-1)-2(3x-2y+18)=0 \\ \text{or,}~~ 6x+9y-3-6x+4y-36=0 \\ \text{or,}~~ 13y-39=0 \\ \therefore~ y=\frac{39}{13}=3.$

Putting the value of $~y~$ in $~(1)~$ we get,

$~2x+3 \times 3=1 \Rightarrow x=\frac{-8}{2}=-4.$

So, the point of intersection of $~(1)~$ and $~(3)~$ is $~B(-4,3).$

If $~C(a,b)~$ is the image of the point $~A(-2,6)~$, then clearly $~B(-4,3)~$ is the midpoint of the straight line $~AB.$

From the picture, we have $~A \equiv (-2,6),~B \equiv (-4,3),~ C \equiv(a,b)~~\text{(say)}$

$ \therefore~\frac{-2+a}{2}=-4 \Rightarrow a=-6,\\~~\frac{6+b}{2}=3 \Rightarrow b=0.$

$~\therefore~ C\equiv (-6,0).$

Hence,  the co-ordinates of the point which is the image of $~(-2,6)~$ with respect to the given line are $~(-6,0).$

10. Two sides of a square have the equations $~5x+12y-10=0~$ and $~5x+12y+29=0.~$ The co-ordinates of a point on the third side are $~(3,5).~$ Find the equations to the other sides of the square.

Solution.

Let $~AB \equiv 5x+12y-10=0\rightarrow(1),\\~CD \equiv 5x+12y+29=0\rightarrow(2).$

Equation of any straight line perpendicular to the straight line $~(1)~$ can be written as $~12x-5y+k=0~(k \neq 0) \rightarrow(3)~$ 

Since the straight line $~(3)~$passes through the point $~(3,5)~$,

$~12 \times 3-5 \times 5+k=0 \\ \text{or,}~~36-25+k=0 \\ \text{or,}~~11+k=0 \\ \therefore~ k=-11.$

$\therefore~ BC \equiv 12x-5y-11=0.$

Again, since $~AD~ ||~BC,~$ the equation of $~AD~$ can be written as $~12x-5y+k'=0~~$ where $~AD=CD.$

$\therefore~\left|\frac{k'+11}{\sqrt{12^2+5^2}}\right|=\left|\frac{29+10}{\sqrt{12^2+5^2}}\right| \\ \text{or,}~~ k'+11=\pm 39 \\ \text{or,}~~k'=39-11=28,\\~~k'=-39-11=-50.$

Hence, $~AD \equiv 12x-5y+28=0~~$ or, $~AD \equiv 12x-5y-50=0.$

So, the equations to the other sides of the square are $~12x-5y+28=0~,~12x-5y-50=0.$

11. Show that the four sls, $~x\cos\alpha+y\sin\alpha=p,~x\sin\alpha-y\cos\alpha=-p,~x\cos\alpha+y\sin\alpha=-p~$ and $~x\sin\alpha-y\cos\alpha=p~$ form a square.

Solution.

Given four straight lines  are as follows :

$~x\cos\alpha+y\sin\alpha=p \\ \text{or,}~~ y=-x\cot\alpha+p\csc \alpha\rightarrow(1),\\~~\\~x\sin\alpha-y\cos\alpha=-p \\ \text{or,}~~ y=x\tan\alpha+p\sec\alpha\rightarrow(2),\\~~\\~x \cos\alpha +y\sin\alpha=-p \\ \text{or,}~~ y=-x\cot\alpha-p\csc\alpha \rightarrow(3)\\~~ \\~x \sin\alpha-y\cos\alpha=p \\ \text{or,}~~ y=x\tan\alpha-p\sec\alpha \rightarrow(4)$

Again, $~\tan\alpha \times (-\cot \alpha)=-1~$ and so,

$\text{slope of (1)} \times \text{slope of (2)}=-1~ $ and $\text{slope of (3)} \times \text{slope of (4)}=-1~ $ .

Hence, the straight lines $~(1),~(2)~$ and $~(3),~(4)~$ are perpendicular to each other.

Again, the straight lines $~(1),~(3)~$ and $~(2),~(4)~$ are parallel to each other.

The distance between the straight lines $~(1)~$ and $~(3)~$ is 

$~\frac{|p\csc\alpha-(-p\csc\alpha)|}{\sqrt{1+(-\cot\alpha)^2}}=\frac{|2p\csc\alpha|}{|\csc\alpha|}=2|p|.$

The distance between the straight lines $~(2)~$ and $~(4)~$ is 

$~\frac{|p\sec\alpha-(-p\sec\alpha)|}{\sqrt{1+\tan^2\alpha}}=\frac{|2p\sec\alpha|}{|\sec\alpha|}=2|p|.$

Since the opposite sides of the given quadrilateral are equal and parallel to each other and so, we can conclude that the given four straight lines form a square.