Straight Line | Part-2 |Ex-2A Examhoop

In the previous article, we discussed 9 short ans type questions of Short Answer Type Questions of Straight Line Chapter of Chhaya Mathematics, Class 11. In this article, we have solved few more.

Short Answer Type Questions of Straight Line Chapter- Ex 2A | S N De

10.The perimeter of the triangle formed by the straight line $~4x+3y-k=0~$ with the co-ordinate axes is $~24~$ unit ; find the value of $~k.$

Solution.

We have the equation of straight line

$(AB) :~4x+3y-k=0 \\ ~~\text{or,}~~ \frac{x}{k/4}+\frac{y}{k/3}=1\rightarrow(1)$

$\therefore~$ the straight line intersects $~x-$ axis at $~A(a,0)~$ and $~y-$ axis at $~B(0,b)~$ making $~\Delta OAB~$ with origin at $~O(0,0).$

By the intercept form of Straight line $~(1)~$ , we see $~OA=\frac k4,~~OB=\frac k3.$

$\therefore~ AB=\sqrt{OA^2+OB^2}=\sqrt{\frac{k^2}{16}+\frac{k^2}{9}} \\ \text{or,}~~AB=\sqrt{\frac{9k^2+16k^2}{144}}=\sqrt{\frac{25k^2}{144}}=\frac{5k}{12}.$

So, the perimeter of the triangle formed by the given straight line

$\frac{k}{4}+\frac k3+\frac{5k}{12}=24 \\ \text{or,}~~ \frac{3k+4k+5k}{12}=24 \\ \text{or,}~~ \frac{12k}{12}=24 \\ \therefore k=24.$

11.If $~a+b+c=0~$ for all positions of the moving line $~ax+by+c=0~,$ Show that the line always passes through a fixed point. Find the co-ordinates of that fixed point.

Solution.

We have

$~a+b+c=0 \Rightarrow c=-a-b \rightarrow(1)$

$\text{Now,}~~ ax+by+c=0 \\ \text{or,}~~ ax+by-a-b=0 ~~[\text{By (1)}] \\ \text{or,}~~a(x-1)+b(y-1)=0 \\ \text{or,}~~(x-1)+\frac ba(y-1)=0 ~~ [~a \neq 0] \\ \text{or,}~~ (x-1)+\lambda (y-1)=0 \rightarrow(2) ~~ [\text{where}~~\lambda \neq 0]$

Clearly, $~(2)~$ represents a straight line through the point of intersection of $~x-1=0,~~y-1=0.$ Solving $~x-1=0,~y-1=0~$ we get $~x=1,~y=1~$ and so, the required fixed point is $~(1,1).$

12.Show that the straight line $~(a+2b)x+(a-3b)y +b-a=0~$ always passes through a fixed point; find the co-ordinates of that fixed point.

Solution.

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

Clearly, for different values of $~a,b~$ where $~a~$ and $~b~$ are not simultaneously zero, $~(1)~$ represents straight lines through the point of intersection of the straight lines $~x+y-1=0\rightarrow(2)~$ and $~2x-3y+1=0\rightarrow(3).$

Now, solving $~(2)~$ and $~(3)~$ we get, $~x=\frac 25,~~y=\frac 35.$

So, the co-ordinates of that fixed point is $~\left(\frac 25,\frac 35\right).$

13. Show that the equation of the straight line $~x\cos\alpha+y\sin\alpha=p~$ can be expressed in the following form :

$~\frac{x-p\cos\alpha}{-\sin\alpha}=\frac{y-p\sin\alpha}{\cos\alpha}=r.$

Solution.

$x\cos\alpha+y\sin\alpha=p \\ \text{or,}~~ x\cos\alpha+y\sin\alpha=p(\cos^2\alpha+\sin^2\alpha) \\ \text{or,}~~ \cos\alpha(x-p\cos\alpha)=-\sin\alpha(y-p\sin\alpha) \\ \text{or,}~~ \frac{x-p\cos\alpha}{-\sin\alpha}=\frac{y-p\sin\alpha}{\cos\alpha}=r,$

$\text{where}~~ r=\frac{\sqrt{(x-p\cos\alpha)^2+(y-p\sin\alpha)^2}}{\sqrt{\sin^2\alpha+\cos^2\alpha}}=\sqrt{(x-p\cos\alpha)^2+(y-p\sin\alpha)^2}$

14.If $~p_1~$ and $~p_2~$ be the length of the perpendiculars from the origin upon the lines $~4x+3y=5\cos\alpha~$ and $~6x-8y=5\sin\alpha~$ respectively, show that $~p_1^2+4p_2^2=1.$

Solution.

$p_1=\left|\frac{4 \times 0+3 \times 0-5 \cos\alpha}{\sqrt{4^2+3^2}}\right|=\frac{5|\cos\alpha|}{5} \\~~ \text{or,}~~|\cos\alpha|=p_1 \rightarrow(1)$

$p_2=\left|\frac{6 \times 0-8\times 0-5\sin\alpha}{\sqrt{6^2+8^2}}\right|=\frac{5|\sin\alpha|}{10} \\~~ \text{or,}~~ |\sin\alpha|=2p_2\rightarrow(2)$

Hence, by $~(1)~$ and $~(2)~$ we get,

$\sin^2\alpha+\cos^2\alpha=1 \\ ~~\text{or,}~~ (2p_2)^2+p_1^2=1 \\ ~~ \text{or,}~~ p_1^2+4p_2^2=1.$

15. Find the equation of the straight line through the point $~(3,2)~$ and the point of intersection of the lines $~3x+y-5=0~$ and $~x+5y+3=0.~$ Also find the length of the portion of the line intercepted between the co-ordinate axes.

Solution.

The equation of the straight line through the point of intersection of the straight lines $~3x+y-5=0~$ and $~x+5y+3=0~$ is

$3x+y-5+k(x+5y+3)=0 \\ \text{or,}~~ (3+k)x+(1+5k)y-5+3k=0\rightarrow(1)$

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

$(3+k)\times 3+(1+5k) \times 2-5+3k=0 \\ \text{or,}~~ 16k=-6 \\ \text{or,}~~ k=-\frac{3}{8}.$

Hence, putting the value of $~k~$ in $~(1),~$ we get

$(3-3/8)x+\left(1-\frac{15}{8}\right)y-5+3(-3/8)=0 \\ \text{or,}~~ 21x-7y-49=0 \\ \text{or,}~~ 3x-y=7 \\ \text{or,}~~ \frac{x}{7/3}+\frac{y}{-7}=1\rightarrow(2)$

So, $~(2)~$ represents the intercept form of straight line that cuts $~x-$ axis at $~A(7/3,0)~$ and $~y-$ axis at $~B(0,-7).$

So, $~AB=\sqrt{(7/3-0)^2+(0+7)^2}=7\sqrt{\frac 19+1}=\frac 73 \sqrt{10}~\text{unit.}$

16.If the straight line $~ax+y+6=0~$ passes through the point of intersection of the lines $~x+y+4=0~$ and $~2x+3y+10=0,~$ find $~a.$

Solution.

The equation of the straight line through the point of intersection of the straight lines $~x+y+4=0~$ and $~2x+3y+10=0~$ is

$x+y+4+k(2x+3y+10)=0 ~~(k \neq 0) \\ \text{or,}~~ (1+2k)x+(1+3k)y+4+10k=0 \rightarrow(1)$

Now the straight line $~(1)~$ and $~ax+y+6=0~$ are identical.

So, $~\frac{a}{1+2k}=\frac{1}{1+3k}=\frac{6}{4+10k}$

$\therefore~\frac{1}{1+3k}=\frac{6}{4+10k} \\~~ \text{or,}~~ 4+10k=6+18k \\ ~~\text{or,}~~ 8k=-2 \\ ~~\text{or,}~~ k=-\frac 14$

Again,

$\frac{a}{1+2k}=\frac{1}{1+3k} \\ \text{or,}~~ \frac{a}{1+2 \times (-1/4)}=\frac{1}{1+3 \times (-1/4)} \\ \text{or,}~~ \frac{a}{1-\frac 12}=\frac{1}{1-\frac 34} \\ \text{or,}~~ \frac{a}{1/2}=\frac{1}{1/4} \\ \text{or,}~~ 2a=4 \\ \text{or,}~~ a=2.$

17.Find the equation of the straight line which passes through the point of intersection of the straight lines $~3x-4y+1=0~$ and $~5x+y-1=0~$ and makes equal intercepts upon the co-ordinate axes.

Solution.

18.Find the equation of the straight line which makes intercepts on the axes equal in magnitude but opposite in sign and passes through the point of intersection of the lines $~x+3y+4=0~$ and $~2x-y=13.~$ Also find the perpendicular distance of the line from the origin.

Solution.

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Short Answer Type Questions of Straight Line Chapter- Ex 2A | S N De

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