Differentiation (Part-38) | S N De

 

$10.~~$ A straight line passes through the point $~A(1,2)~$ and makes an angle $~\theta~$ with the positive direction of $~x-$axis. Find $~\theta~$, given that the distance of $~A~$ from the point of intersection of this line with the line $~x+y=4~$ is $~\frac{\sqrt{6}}{3}~$unit.

Solution.

Equation of any straight line (with slope $~m~$) passing through the point $~A(1,2)~$ can be written as $~y-2=m(x-1)\\ \Rightarrow mx-y=m-2 \rightarrow(1)$

Again, $~x+y=4 \rightarrow(2)$

Adding $~(1)~$ and $~(2)~$ we get, 

$~(mx-y)+(x+y)=(m-2)+4 \\ \text{or,}~~ (m+1)x=m+2 \\ \text{or,}~~ x=\frac{m+2}{m+1}.$

So, by $~(2)$

$y=4-x\\=4-\left(\frac{m+2}{m+1}\right)\\=\frac{4(m+1)-(m+2)}{m+1}\\=\frac{3m+2}{m+1}$

The distance$~(d)~$ between the points $~(1,2)~$ and $~\left(\frac{m+2}{m+1}\right)~$ is

$d=\sqrt{\left(\frac{m+2}{m+1}-1\right)^2+\left(\frac{3m+2}{m+1}-2\right)^2}\\ \text{or,}~~d^2=\left(\frac{m+2-m-1}{m+1}\right)^2+\left(\frac{3m+2-2m-2}{m+1}\right)^2 \\ \text{or,}~~\left(\frac{\sqrt{6}}{3}\right)^2=\frac{1}{(m+1)^2}(1+m^2)\\ \text{or,}~~\frac 69=\frac 23=\frac{m^2+1}{(m+1)^2} \\ \text{or,}~~ 2(m+1)^2=3(m^2+1) \\ \text{or,}~~2(m^2+2m+1)=3m^2+3 \\ \text{or,}~~m^2-4m+1=0 \\ \therefore~ m=\frac{4 \pm \sqrt{(-4)^2-4 \cdot 1 \cdot 1}}{2 \cdot 1} \\ \text{or,}~~ m=\frac{4 +2\sqrt{3}}{2}~~[\text{Ignoring -ve sign}] \\ \text{or,}~~ m=\tan\theta=2+\sqrt{3} \\ \text{or,}~~\tan\theta=\tan 75^{\circ} \\ \therefore~\theta =75^{\circ}.$

$11.~~$ Find the equations to the straight lines which are at a distance of $~1~$ unit from the origin and which pass through the point $~(3,1).$

Solution.

The equation of straight line passing through $~(3,1)~$ and having slope $~m~$ is 

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

By question, 

$~\frac{|1-3m|}{\sqrt{m^2+1}}=1 \\ \text{or,}~~ \frac{1-3m}{\sqrt{m^2+1}}=\pm 1 \\ \text{or,}~~ (1-3m)^2=m^2+1 \\ \text{or,}~~ 1-6m+9m^2=m^2+1 \\ \text{or,}~~ 8m^2-6m=0 \\ \text{or,}~~ 2m(4m-3)=0 \\ \therefore~ m=0,~\frac 34.$

For $~m=0,~$ the equation of the straight line is :

 $~y-1=0~~\text{[By (1)]}$

For $~m=\frac 34,~$ the equation of the straight line is :

 $~y-1=\frac 34(x-3)~~\text{[By (1)]} \\ \text{or,}~~ 3x-4y=5.$ 

$12.~~$ Show that the line joining the origin and $~(2,3)~$ is concurrent with the lines $~5x-3y-2=0~$ and $~x+y=10.$

Solution.

The point of intersection of the lines $~5x-3y-2=0\rightarrow(1)~$ and $~x+y=10\rightarrow(2)~$ is $~(4,6)~$ which can be obtained by solving $~(1)~$ and $~(2).$

The equation of the straight line joining $~(0,0)~$ and $~(2,3)~$ is given by : 

$~y-0=\frac 32 (x-0) \\ \text{or,}~~ 3x-2y=0\rightarrow(3)$

Now, the point $~(4,6)~$ satisfies the equation $~(3)$ and so we can finally conclude that the line joining the origin and $~(2,3)~$ is concurrent with the lines $~5x-3y-2=0~$ and $~x+y=10.$

$13.~~$ Prove that the straight lines $~ax+(b+c)y+d=0,~bx+(c+a)y+d=0~$  and $~cx+(a+b)y+d=0~$ are concurrent.

Solution.

 We have,  $~ax+(b+c)y+d=0 \rightarrow(1),~bx+(c+a)y+d=0\rightarrow(2)~$  and $~cx+(a+b)y+d=0~\rightarrow(3).$

From $~(1)~$ and $~(2)~$ we get,

$\frac{x}{(b+c)d-(c+a)d}=\frac{y}{bd-ad}=\frac{1}{a(c+a)-b(b+c)} \\ \Rightarrow \frac{x}{d(b+c-c-a)}=\frac{y}{d(b-a)}=\frac{1}{ac+a^2-b^2-bc} \\ \Rightarrow \frac{x}{(b-a)d}=\frac{y}{d(b-a)}=\frac{1}{(a^2-b^2)+c(a-b)} \\ \Rightarrow \frac{x}{(b-a)d}=\frac{y}{d(b-a)}=\frac{1}{(a+b)(a-b)+c(a-b)} \\ \Rightarrow \frac{x}{(b-a)d}=\frac{y}{d(b-a)}=\frac{1}{(a-b)(a+b+c)} \\ \therefore~x=\frac{-d}{a+b+c},~~y=\frac{-d}{a+b+c}.$

Now, the point $~\left(\frac{-d}{a+b+c},~~\frac{-d}{a+b+c}\right)~$ satisfies the straight line $~(3)~$ and so, we can conclude that the straight lines $~ax+(b+c)y+d=0,~bx+(c+a)y+d=0~$  and $~cx+(a+b)y+d=0~$ are concurrent.

$14.~~$ Find the condition so that the lines $~x\cos\alpha+y\sin\alpha=p,~x\cos\beta+y\sin\beta=q~$ and $~y=x\tan\theta~$ be concurrent.

Solution.

We have $~x\cos\alpha+y\sin\alpha=p \rightarrow(1),~x\cos\beta+y\sin\beta=q\rightarrow(2)~$

From $~(1)~$ and $~(2)~$ we get by cross-multiplication,

$\frac{x}{-q\sin\alpha+p\sin\beta}=\frac{y}{-p\cos\beta+q\cos\alpha}=\frac{1}{\cos\alpha\sin\beta-\sin\alpha\cos\beta} \\ \Rightarrow \frac{x}{p\sin\beta-q\sin\alpha}=\frac{y}{q\cos\alpha-p\cos\beta}=\frac{1}{\sin(\beta-\alpha)} \\ \therefore x=\frac{p\sin\beta-q\sin\alpha}{\sin(\beta-\alpha)},~y=\frac{q\cos\alpha-p\cos\beta}{\sin(\beta-\alpha)}.$

If the given lines are to be concurrent then, the point

 $~\left(\frac{p\sin\beta-q\sin\alpha}{\sin(\beta-\alpha)},~\frac{q\cos\alpha-p\cos\beta}{\sin(\beta-\alpha)}\right)~$ must satisfy the third straight line. 

So, $~\frac{q\cos\alpha-p\cos\beta}{\sin(\beta-\alpha)}=\frac{p\sin\beta-q\sin\alpha}{\sin(\beta-\alpha)} \cdot \tan\theta \\ \text{or,}~~p\cos(\theta-\beta)=q\cos(\theta-\alpha).$

$15.~~$ If $~ab+bc+ca=0,~$ show that the lines $~\frac xa+\frac yb=\frac 1c,~\frac xb+\frac yc=\frac 1a~$ and $~\frac xc+\frac ya=\frac 1b~$ are concurrent.

Solution.

 $~ab+bc+ca=0\\ \Rightarrow \frac 1a+\frac 1b+\frac 1c=1\rightarrow(1)$

Using $~(1),~$ the given equation of straight lines can be written as

$~\frac xa+\frac yb=-\frac 1a-\frac 1b,\\~\frac xb+\frac yc=-\frac 1b-\frac 1c,\\~\frac xc+\frac ya=-\frac 1c-\frac 1a.$

From the aforesaid equations of straight lines, we can say that all straight lines pass through the point $~(-1,-1)~$ and that means the straight lines are concurrent.

$16.~~$ The variable coefficients $~p, q, r~$ in the equation of the straight line $~px+qy+r=0~$ are connected by the relation $~pa+qb+rc=0~$ where $~a,b,c~$ are fixed constants. Show that the variable line always passes through a fixed point.

Solution.

$~pa+qb+rc=0 \\ \Rightarrow \frac{pa}{c}+\frac{qb}{c}+r=0\rightarrow(1)$

From $~(1),~$ we can say that given equation $~px+qy+r=0~$ is satisfied by $~x=\frac ac,~y=\frac bc~$.

Clearly, the straight line $~px+qy+r=0~$ passes through the point $~\left(\frac ac,\frac bc\right)~$ which is a fixed point.

$17.~~$ Show that the area of the triangle formed by the straight lines $~y=m_1x+c_1,~y=m_2x+c_2~$ and $~x=0~$ is $~\frac 12\frac{(c_1-c_2)^2}{|m_1-m_2|}.$

Solution.

We have, $~y=m_1x+c_1 \rightarrow(1),~y=m_2x+c_2\rightarrow(2),~x=0\rightarrow(3).$

From $~(1),~(2),~(3)~$ we get, $~A(0,c_1),~C(0,c_2),~B\left(\frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2}\right).$

Area of $~\Delta ABC~$ is 

$=\frac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\\=\frac 12\left|0+0+\frac{c_2-c_1}{m_1-m_2} \cdot (c_1-c_2)\right|\\=\frac 12 \cdot \frac{(c_1-c_2)^2}{|m_1-m_2|}.$

$18.~~$ A variable straight line through the point of intersection of the two lines $~\frac x3+\frac y2=1~$ and $~\frac x2+\frac y3=1~$ meets the co-ordinate axes at $~ A~$ and $~B~$. Find the locus of the middle point of $~AB.$

Solution.

Any straight line through the point of intersection of two given lines can be written as 

$~\frac x3+\frac y2-1+k\left(\frac x2+\frac y3-1\right)=0 \\ \text{or,}~~ \frac{x}{\frac{6(k+1)}{2+3k}}+\frac{y}{\frac{6(k+1)}{3+2k}}=1$

So, $~A \equiv \left(\frac{6(k+1)}{2+3k},0\right),~B\equiv \left(0,\frac{6(k+1)}{3+2k}\right).$

If $~(\alpha,\beta)~$ is the middle point of $~\overline{AB}~$ then

$~\alpha=\frac{\frac{6k+6}{2+3k}+0}{2} \rightarrow(1),\\~\beta=\frac{\frac{6k+6}{3+2k}+0}{2}\rightarrow(2).$

By $~(1),~$ we get

$~2\alpha=\frac{6k+6}{2+3k} \\ \text{or,}~~ \alpha=\frac{3k+3}{2+3k} \\ \text{or,}~~  2\alpha+3\alpha k=3k+3 \\ \text{or,}~~  k(3\alpha-3)=3-2\alpha \\ \text{or,}~~  k=\frac{3-2\alpha}{3\alpha-3}\rightarrow(3).$

Similarly, by $~(2)~$ we get,

$~k=\frac{3-3\beta}{2\beta-3}\rightarrow(4)$

From $~(3),~(4)~$ we get ,

$~\frac{3-2\alpha}{3\alpha-3}=\frac{3-3\beta}{2\beta-3} \\ \text{or,}~~ 5\alpha\beta=3(\alpha+\beta).$

So,   the locus of the middle point of $~AB~$ is given by $~5xy=3(x+y).$